Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHExIntrod_Zan97

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

Q DP problem:
The TRS P consists of the following rules:

P₁(mark₁(X)) -> P₁(X)
ACTIVE₁(zero₁(X)) -> ZERO₁(active₁(X))
PROPER₁(prod₂(X1, X2)) -> PROPER₁(X2)
S₁(mark₁(X)) -> S₁(X)
ACTIVE₁(p₁(X)) -> ACTIVE₁(X)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROD₂(mark₁(X1), X2) -> PROD₂(X1, X2)
ZERO₁(mark₁(X)) -> ZERO₁(X)
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
PROD₂(ok₁(X1), ok₁(X2)) -> PROD₂(X1, X2)
ACTIVE₁(prod₂(X1, X2)) -> ACTIVE₁(X2)
FACT₁(mark₁(X)) -> FACT₁(X)
PROPER₁(zero₁(X)) -> PROPER₁(X)
P₁(ok₁(X)) -> P₁(X)
ACTIVE₁(add₂(X1, X2)) -> ADD₂(active₁(X1), X2)
ACTIVE₁(fact₁(X)) -> FACT₁(p₁(X))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(prod₂(s₁(X), Y)) -> ADD₂(Y, prod₂(X, Y))
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
ACTIVE₁(add₂(X1, X2)) -> ADD₂(X1, active₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(fact₁(X)) -> PROD₂(X, fact₁(p₁(X)))
ACTIVE₁(add₂(s₁(X), Y)) -> ADD₂(X, Y)
PROPER₁(prod₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(p₁(X)) -> PROPER₁(X)
PROPER₁(prod₂(X1, X2)) -> PROD₂(proper₁(X1), proper₁(X2))
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(fact₁(X)) -> FACT₁(active₁(X))
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)
ACTIVE₁(fact₁(X)) -> P₁(X)
PROD₂(X1, mark₁(X2)) -> PROD₂(X1, X2)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(add₂(X1, X2)) -> ADD₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
PROPER₁(fact₁(X)) -> PROPER₁(X)
PROPER₁(p₁(X)) -> P₁(proper₁(X))
ACTIVE₁(prod₂(X1, X2)) -> PROD₂(active₁(X1), X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
FACT₁(ok₁(X)) -> FACT₁(X)
PROPER₁(fact₁(X)) -> FACT₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(prod₂(s₁(X), Y)) -> PROD₂(X, Y)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)
ACTIVE₁(prod₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(fact₁(X)) -> S₁(0)
ACTIVE₁(zero₁(X)) -> ACTIVE₁(X)
ACTIVE₁(fact₁(X)) -> IF₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X))))
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(p₁(X)) -> P₁(active₁(X))
ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)
PROPER₁(zero₁(X)) -> ZERO₁(proper₁(X))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)
ACTIVE₁(prod₂(X1, X2)) -> PROD₂(X1, active₁(X2))
ACTIVE₁(fact₁(X)) -> ZERO₁(X)
ZERO₁(ok₁(X)) -> ZERO₁(X)
ACTIVE₁(add₂(s₁(X), Y)) -> S₁(add₂(X, Y))

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

P₁(mark₁(X)) -> P₁(X)
ACTIVE₁(zero₁(X)) -> ZERO₁(active₁(X))
PROPER₁(prod₂(X1, X2)) -> PROPER₁(X2)
S₁(mark₁(X)) -> S₁(X)
ACTIVE₁(p₁(X)) -> ACTIVE₁(X)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROD₂(mark₁(X1), X2) -> PROD₂(X1, X2)
ZERO₁(mark₁(X)) -> ZERO₁(X)
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
PROD₂(ok₁(X1), ok₁(X2)) -> PROD₂(X1, X2)
ACTIVE₁(prod₂(X1, X2)) -> ACTIVE₁(X2)
FACT₁(mark₁(X)) -> FACT₁(X)
PROPER₁(zero₁(X)) -> PROPER₁(X)
P₁(ok₁(X)) -> P₁(X)
ACTIVE₁(add₂(X1, X2)) -> ADD₂(active₁(X1), X2)
ACTIVE₁(fact₁(X)) -> FACT₁(p₁(X))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(prod₂(s₁(X), Y)) -> ADD₂(Y, prod₂(X, Y))
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
ACTIVE₁(add₂(X1, X2)) -> ADD₂(X1, active₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(fact₁(X)) -> PROD₂(X, fact₁(p₁(X)))
ACTIVE₁(add₂(s₁(X), Y)) -> ADD₂(X, Y)
PROPER₁(prod₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(p₁(X)) -> PROPER₁(X)
PROPER₁(prod₂(X1, X2)) -> PROD₂(proper₁(X1), proper₁(X2))
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(fact₁(X)) -> FACT₁(active₁(X))
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)
ACTIVE₁(fact₁(X)) -> P₁(X)
PROD₂(X1, mark₁(X2)) -> PROD₂(X1, X2)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(add₂(X1, X2)) -> ADD₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
PROPER₁(fact₁(X)) -> PROPER₁(X)
PROPER₁(p₁(X)) -> P₁(proper₁(X))
ACTIVE₁(prod₂(X1, X2)) -> PROD₂(active₁(X1), X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
FACT₁(ok₁(X)) -> FACT₁(X)
PROPER₁(fact₁(X)) -> FACT₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(prod₂(s₁(X), Y)) -> PROD₂(X, Y)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)
ACTIVE₁(prod₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(fact₁(X)) -> S₁(0)
ACTIVE₁(zero₁(X)) -> ACTIVE₁(X)
ACTIVE₁(fact₁(X)) -> IF₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X))))
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(p₁(X)) -> P₁(active₁(X))
ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)
PROPER₁(zero₁(X)) -> ZERO₁(proper₁(X))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)
ACTIVE₁(prod₂(X1, X2)) -> PROD₂(X1, active₁(X2))
ACTIVE₁(fact₁(X)) -> ZERO₁(X)
ZERO₁(ok₁(X)) -> ZERO₁(X)
ACTIVE₁(add₂(s₁(X), Y)) -> S₁(add₂(X, Y))

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 10 SCCs with 28 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)
ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD₂(ok₁(X1), ok₁(X2)) -> ADD₂(X1, X2)

Used argument filtering: ADD₂(x1, x2) = x2
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)
ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD₂(X1, mark₁(X2)) -> ADD₂(X1, X2)

Used argument filtering: ADD₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD₂(mark₁(X1), X2) -> ADD₂(X1, X2)

Used argument filtering: ADD₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

P₁(mark₁(X)) -> P₁(X)
P₁(ok₁(X)) -> P₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

P₁(ok₁(X)) -> P₁(X)

Used argument filtering: P₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

P₁(mark₁(X)) -> P₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

P₁(mark₁(X)) -> P₁(X)

Used argument filtering: P₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROD₂(ok₁(X1), ok₁(X2)) -> PROD₂(X1, X2)
PROD₂(X1, mark₁(X2)) -> PROD₂(X1, X2)
PROD₂(mark₁(X1), X2) -> PROD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROD₂(X1, mark₁(X2)) -> PROD₂(X1, X2)

Used argument filtering: PROD₂(x1, x2) = x2
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROD₂(ok₁(X1), ok₁(X2)) -> PROD₂(X1, X2)
PROD₂(mark₁(X1), X2) -> PROD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROD₂(ok₁(X1), ok₁(X2)) -> PROD₂(X1, X2)

Used argument filtering: PROD₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROD₂(mark₁(X1), X2) -> PROD₂(X1, X2)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROD₂(mark₁(X1), X2) -> PROD₂(X1, X2)

Used argument filtering: PROD₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

S₁(ok₁(X)) -> S₁(X)
S₁(mark₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(mark₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

S₁(ok₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(ok₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ZERO₁(mark₁(X)) -> ZERO₁(X)
ZERO₁(ok₁(X)) -> ZERO₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ZERO₁(ok₁(X)) -> ZERO₁(X)

Used argument filtering: ZERO₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ZERO₁(mark₁(X)) -> ZERO₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ZERO₁(mark₁(X)) -> ZERO₁(X)

Used argument filtering: ZERO₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)

Used argument filtering: IF₃(x1, x2, x3) = x3
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)

Used argument filtering: IF₃(x1, x2, x3) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FACT₁(mark₁(X)) -> FACT₁(X)
FACT₁(ok₁(X)) -> FACT₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FACT₁(ok₁(X)) -> FACT₁(X)

Used argument filtering: FACT₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FACT₁(mark₁(X)) -> FACT₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FACT₁(mark₁(X)) -> FACT₁(X)

Used argument filtering: FACT₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(zero₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(fact₁(X)) -> PROPER₁(X)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(prod₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(prod₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(p₁(X)) -> PROPER₁(X)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(add₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(prod₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(prod₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(add₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
add₂(x1, x2) = add₂(x1, x2)
zero₁(x1) = x1
s₁(x1) = x1
fact₁(x1) = x1
if₃(x1, x2, x3) = if₃(x1, x2, x3)
prod₂(x1, x2) = prod₂(x1, x2)
p₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(zero₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(fact₁(X)) -> PROPER₁(X)
PROPER₁(p₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(p₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
zero₁(x1) = x1
s₁(x1) = x1
fact₁(x1) = x1
p₁(x1) = p₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(zero₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(fact₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(fact₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
zero₁(x1) = x1
s₁(x1) = x1
fact₁(x1) = fact₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(zero₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
zero₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER₁(zero₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(zero₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
zero₁(x1) = zero₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(prod₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(prod₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(zero₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(p₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(prod₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(prod₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(add₂(X1, X2)) -> ACTIVE₁(X2)

Used argument filtering: ACTIVE₁(x1) = x1
prod₂(x1, x2) = prod₂(x1, x2)
add₂(x1, x2) = add₂(x1, x2)
fact₁(x1) = x1
if₃(x1, x2, x3) = x1
zero₁(x1) = x1
s₁(x1) = x1
p₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)
ACTIVE₁(zero₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(p₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(p₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
if₃(x1, x2, x3) = x1
fact₁(x1) = x1
zero₁(x1) = x1
s₁(x1) = x1
p₁(x1) = p₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(zero₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
fact₁(x1) = x1
if₃(x1, x2, x3) = x1
zero₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)
ACTIVE₁(zero₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(zero₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
if₃(x1, x2, x3) = x1
fact₁(x1) = x1
zero₁(x1) = zero₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
fact₁(x1) = x1
if₃(x1, x2, x3) = if₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(fact₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
fact₁(x1) = fact₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

                                  ↳ QDP

                                    ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The TRS R consists of the following rules:

active₁(fact₁(X)) -> mark₁(if₃(zero₁(X), s₁(0), prod₂(X, fact₁(p₁(X)))))
active₁(add₂(0, X)) -> mark₁(X)
active₁(add₂(s₁(X), Y)) -> mark₁(s₁(add₂(X, Y)))
active₁(prod₂(0, X)) -> mark₁(0)
active₁(prod₂(s₁(X), Y)) -> mark₁(add₂(Y, prod₂(X, Y)))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(zero₁(0)) -> mark₁(true)
active₁(zero₁(s₁(X))) -> mark₁(false)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(fact₁(X)) -> fact₁(active₁(X))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(zero₁(X)) -> zero₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(prod₂(X1, X2)) -> prod₂(active₁(X1), X2)
active₁(prod₂(X1, X2)) -> prod₂(X1, active₁(X2))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(add₂(X1, X2)) -> add₂(active₁(X1), X2)
active₁(add₂(X1, X2)) -> add₂(X1, active₁(X2))
fact₁(mark₁(X)) -> mark₁(fact₁(X))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
zero₁(mark₁(X)) -> mark₁(zero₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
prod₂(mark₁(X1), X2) -> mark₁(prod₂(X1, X2))
prod₂(X1, mark₁(X2)) -> mark₁(prod₂(X1, X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
add₂(mark₁(X1), X2) -> mark₁(add₂(X1, X2))
add₂(X1, mark₁(X2)) -> mark₁(add₂(X1, X2))
proper₁(fact₁(X)) -> fact₁(proper₁(X))
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(zero₁(X)) -> zero₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(prod₂(X1, X2)) -> prod₂(proper₁(X1), proper₁(X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(add₂(X1, X2)) -> add₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
fact₁(ok₁(X)) -> ok₁(fact₁(X))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
zero₁(ok₁(X)) -> ok₁(zero₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
prod₂(ok₁(X1), ok₁(X2)) -> ok₁(prod₂(X1, X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
add₂(ok₁(X1), ok₁(X2)) -> ok₁(add₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.