Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHEx1_GM03

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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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:

LEQ₂(mark₁(X1), X2) -> LEQ₂(X1, X2)
ACTIVE₁(diff₂(X, Y)) -> P₁(X)
P₁(mark₁(X)) -> P₁(X)
LEQ₂(X1, mark₁(X2)) -> LEQ₂(X1, X2)
PROPER₁(leq₂(X1, X2)) -> LEQ₂(proper₁(X1), proper₁(X2))
ACTIVE₁(diff₂(X1, X2)) -> DIFF₂(active₁(X1), X2)
S₁(mark₁(X)) -> S₁(X)
ACTIVE₁(p₁(X)) -> ACTIVE₁(X)
ACTIVE₁(leq₂(X1, X2)) -> LEQ₂(active₁(X1), X2)
PROPER₁(diff₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
ACTIVE₁(diff₂(X, Y)) -> LEQ₂(X, Y)
ACTIVE₁(diff₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(diff₂(X, Y)) -> S₁(diff₂(p₁(X), Y))
ACTIVE₁(leq₂(s₁(X), s₁(Y))) -> LEQ₂(X, Y)
DIFF₂(mark₁(X1), X2) -> DIFF₂(X1, X2)
P₁(ok₁(X)) -> P₁(X)
ACTIVE₁(leq₂(X1, X2)) -> LEQ₂(X1, active₁(X2))
ACTIVE₁(leq₂(X1, X2)) -> ACTIVE₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(diff₂(X, Y)) -> IF₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y)))
PROPER₁(diff₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(p₁(X)) -> PROPER₁(X)
ACTIVE₁(diff₂(X1, X2)) -> DIFF₂(X1, active₁(X2))
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(leq₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(leq₂(X1, X2)) -> PROPER₁(X2)
DIFF₂(ok₁(X1), ok₁(X2)) -> DIFF₂(X1, X2)
ACTIVE₁(diff₂(X1, X2)) -> ACTIVE₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
PROPER₁(diff₂(X1, X2)) -> DIFF₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
PROPER₁(p₁(X)) -> P₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
S₁(ok₁(X)) -> S₁(X)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
DIFF₂(X1, mark₁(X2)) -> DIFF₂(X1, X2)
ACTIVE₁(diff₂(X, Y)) -> DIFF₂(p₁(X), Y)
ACTIVE₁(leq₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(p₁(X)) -> P₁(active₁(X))
LEQ₂(ok₁(X1), ok₁(X2)) -> LEQ₂(X1, X2)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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:

LEQ₂(mark₁(X1), X2) -> LEQ₂(X1, X2)
ACTIVE₁(diff₂(X, Y)) -> P₁(X)
P₁(mark₁(X)) -> P₁(X)
LEQ₂(X1, mark₁(X2)) -> LEQ₂(X1, X2)
PROPER₁(leq₂(X1, X2)) -> LEQ₂(proper₁(X1), proper₁(X2))
ACTIVE₁(diff₂(X1, X2)) -> DIFF₂(active₁(X1), X2)
S₁(mark₁(X)) -> S₁(X)
ACTIVE₁(p₁(X)) -> ACTIVE₁(X)
ACTIVE₁(leq₂(X1, X2)) -> LEQ₂(active₁(X1), X2)
PROPER₁(diff₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(if₃(X1, X2, X3)) -> IF₃(active₁(X1), X2, X3)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
IF₃(mark₁(X1), X2, X3) -> IF₃(X1, X2, X3)
ACTIVE₁(diff₂(X, Y)) -> LEQ₂(X, Y)
ACTIVE₁(diff₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(diff₂(X, Y)) -> S₁(diff₂(p₁(X), Y))
ACTIVE₁(leq₂(s₁(X), s₁(Y))) -> LEQ₂(X, Y)
DIFF₂(mark₁(X1), X2) -> DIFF₂(X1, X2)
P₁(ok₁(X)) -> P₁(X)
ACTIVE₁(leq₂(X1, X2)) -> LEQ₂(X1, active₁(X2))
ACTIVE₁(leq₂(X1, X2)) -> ACTIVE₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(diff₂(X, Y)) -> IF₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y)))
PROPER₁(diff₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(p₁(X)) -> PROPER₁(X)
ACTIVE₁(diff₂(X1, X2)) -> DIFF₂(X1, active₁(X2))
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(leq₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(leq₂(X1, X2)) -> PROPER₁(X2)
DIFF₂(ok₁(X1), ok₁(X2)) -> DIFF₂(X1, X2)
ACTIVE₁(diff₂(X1, X2)) -> ACTIVE₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
IF₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> IF₃(X1, X2, X3)
PROPER₁(diff₂(X1, X2)) -> DIFF₂(proper₁(X1), proper₁(X2))
PROPER₁(if₃(X1, X2, X3)) -> IF₃(proper₁(X1), proper₁(X2), proper₁(X3))
PROPER₁(p₁(X)) -> P₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
S₁(ok₁(X)) -> S₁(X)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
DIFF₂(X1, mark₁(X2)) -> DIFF₂(X1, X2)
ACTIVE₁(diff₂(X, Y)) -> DIFF₂(p₁(X), Y)
ACTIVE₁(leq₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(p₁(X)) -> P₁(active₁(X))
LEQ₂(ok₁(X1), ok₁(X2)) -> LEQ₂(X1, X2)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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 8 SCCs with 20 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DIFF₂(X1, mark₁(X2)) -> DIFF₂(X1, X2)
DIFF₂(ok₁(X1), ok₁(X2)) -> DIFF₂(X1, X2)
DIFF₂(mark₁(X1), X2) -> DIFF₂(X1, X2)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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.

DIFF₂(ok₁(X1), ok₁(X2)) -> DIFF₂(X1, X2)

Used argument filtering: DIFF₂(x1, x2) = x2
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DIFF₂(X1, mark₁(X2)) -> DIFF₂(X1, X2)
DIFF₂(mark₁(X1), X2) -> DIFF₂(X1, X2)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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.

DIFF₂(X1, mark₁(X2)) -> DIFF₂(X1, X2)

Used argument filtering: DIFF₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DIFF₂(mark₁(X1), X2) -> DIFF₂(X1, X2)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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.

DIFF₂(mark₁(X1), X2) -> DIFF₂(X1, X2)

Used argument filtering: DIFF₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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

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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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

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

LEQ₂(ok₁(X1), ok₁(X2)) -> LEQ₂(X1, X2)
LEQ₂(mark₁(X1), X2) -> LEQ₂(X1, X2)
LEQ₂(X1, mark₁(X2)) -> LEQ₂(X1, X2)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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.

LEQ₂(X1, mark₁(X2)) -> LEQ₂(X1, X2)

Used argument filtering: LEQ₂(x1, x2) = x2
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LEQ₂(ok₁(X1), ok₁(X2)) -> LEQ₂(X1, X2)
LEQ₂(mark₁(X1), X2) -> LEQ₂(X1, X2)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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.

LEQ₂(ok₁(X1), ok₁(X2)) -> LEQ₂(X1, X2)

Used argument filtering: LEQ₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LEQ₂(mark₁(X1), X2) -> LEQ₂(X1, X2)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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.

LEQ₂(mark₁(X1), X2) -> LEQ₂(X1, X2)

Used argument filtering: LEQ₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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

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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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

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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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

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

PROPER₁(diff₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(leq₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(leq₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(diff₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)
PROPER₁(p₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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(diff₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(leq₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(leq₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(diff₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(if₃(X1, X2, X3)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
diff₂(x1, x2) = diff₂(x1, x2)
leq₂(x1, x2) = leq₂(x1, x2)
s₁(x1) = x1
if₃(x1, x2, x3) = if₃(x1, x2, x3)
p₁(x1) = x1
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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
s₁(x1) = x1
p₁(x1) = p₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(s₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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

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

ACTIVE₁(diff₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(if₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(diff₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(leq₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(leq₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(p₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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(diff₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(diff₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(leq₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(leq₂(X1, X2)) -> ACTIVE₁(X2)

Used argument filtering: ACTIVE₁(x1) = x1
diff₂(x1, x2) = diff₂(x1, x2)
if₃(x1, x2, x3) = x1
leq₂(x1, x2) = leq₂(x1, x2)
s₁(x1) = x1
p₁(x1) = x1
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(p₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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
s₁(x1) = x1
p₁(x1) = p₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

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

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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
if₃(x1, x2, x3) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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)

The TRS R consists of the following rules:

active₁(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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
if₃(x1, x2, x3) = if₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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

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(p₁(0)) -> mark₁(0)
active₁(p₁(s₁(X))) -> mark₁(X)
active₁(leq₂(0, Y)) -> mark₁(true)
active₁(leq₂(s₁(X), 0)) -> mark₁(false)
active₁(leq₂(s₁(X), s₁(Y))) -> mark₁(leq₂(X, Y))
active₁(if₃(true, X, Y)) -> mark₁(X)
active₁(if₃(false, X, Y)) -> mark₁(Y)
active₁(diff₂(X, Y)) -> mark₁(if₃(leq₂(X, Y), 0, s₁(diff₂(p₁(X), Y))))
active₁(p₁(X)) -> p₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
active₁(leq₂(X1, X2)) -> leq₂(active₁(X1), X2)
active₁(leq₂(X1, X2)) -> leq₂(X1, active₁(X2))
active₁(if₃(X1, X2, X3)) -> if₃(active₁(X1), X2, X3)
active₁(diff₂(X1, X2)) -> diff₂(active₁(X1), X2)
active₁(diff₂(X1, X2)) -> diff₂(X1, active₁(X2))
p₁(mark₁(X)) -> mark₁(p₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
leq₂(mark₁(X1), X2) -> mark₁(leq₂(X1, X2))
leq₂(X1, mark₁(X2)) -> mark₁(leq₂(X1, X2))
if₃(mark₁(X1), X2, X3) -> mark₁(if₃(X1, X2, X3))
diff₂(mark₁(X1), X2) -> mark₁(diff₂(X1, X2))
diff₂(X1, mark₁(X2)) -> mark₁(diff₂(X1, X2))
proper₁(p₁(X)) -> p₁(proper₁(X))
proper₁(0) -> ok₁(0)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(leq₂(X1, X2)) -> leq₂(proper₁(X1), proper₁(X2))
proper₁(true) -> ok₁(true)
proper₁(false) -> ok₁(false)
proper₁(if₃(X1, X2, X3)) -> if₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(diff₂(X1, X2)) -> diff₂(proper₁(X1), proper₁(X2))
p₁(ok₁(X)) -> ok₁(p₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
leq₂(ok₁(X1), ok₁(X2)) -> ok₁(leq₂(X1, X2))
if₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(if₃(X1, X2, X3))
diff₂(ok₁(X1), ok₁(X2)) -> ok₁(diff₂(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.