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.