Termination w.r.t. Q proof of TRS/TRCSREx1_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:

a__p₁(0) -> 0
a__p₁(s₁(X)) -> mark₁(X)
a__leq₂(0, Y) -> true
a__leq₂(s₁(X), 0) -> false
a__leq₂(s₁(X), s₁(Y)) -> a__leq₂(mark₁(X), mark₁(Y))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__diff₂(X, Y) -> a__if₃(a__leq₂(mark₁(X), mark₁(Y)), 0, s₁(diff₂(p₁(X), Y)))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(leq₂(X1, X2)) -> a__leq₂(mark₁(X1), mark₁(X2))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(diff₂(X1, X2)) -> a__diff₂(mark₁(X1), mark₁(X2))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__p₁(X) -> p₁(X)
a__leq₂(X1, X2) -> leq₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__diff₂(X1, X2) -> diff₂(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__p₁(0) -> 0
a__p₁(s₁(X)) -> mark₁(X)
a__leq₂(0, Y) -> true
a__leq₂(s₁(X), 0) -> false
a__leq₂(s₁(X), s₁(Y)) -> a__leq₂(mark₁(X), mark₁(Y))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__diff₂(X, Y) -> a__if₃(a__leq₂(mark₁(X), mark₁(Y)), 0, s₁(diff₂(p₁(X), Y)))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(leq₂(X1, X2)) -> a__leq₂(mark₁(X1), mark₁(X2))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(diff₂(X1, X2)) -> a__diff₂(mark₁(X1), mark₁(X2))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__p₁(X) -> p₁(X)
a__leq₂(X1, X2) -> leq₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__diff₂(X1, X2) -> diff₂(X1, X2)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK₁(diff₂(X1, X2)) -> MARK₁(X2)
MARK₁(diff₂(X1, X2)) -> A__DIFF₂(mark₁(X1), mark₁(X2))
A__LEQ₂(s₁(X), s₁(Y)) -> MARK₁(X)
A__DIFF₂(X, Y) -> A__IF₃(a__leq₂(mark₁(X), mark₁(Y)), 0, s₁(diff₂(p₁(X), Y)))
A__LEQ₂(s₁(X), s₁(Y)) -> A__LEQ₂(mark₁(X), mark₁(Y))
MARK₁(diff₂(X1, X2)) -> MARK₁(X1)
A__IF₃(false, X, Y) -> MARK₁(Y)
A__DIFF₂(X, Y) -> A__LEQ₂(mark₁(X), mark₁(Y))
MARK₁(p₁(X)) -> A__P₁(mark₁(X))
MARK₁(leq₂(X1, X2)) -> A__LEQ₂(mark₁(X1), mark₁(X2))
A__DIFF₂(X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__DIFF₂(X, Y) -> MARK₁(Y)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(leq₂(X1, X2)) -> MARK₁(X2)
MARK₁(leq₂(X1, X2)) -> MARK₁(X1)
A__LEQ₂(s₁(X), s₁(Y)) -> MARK₁(Y)
A__IF₃(true, X, Y) -> MARK₁(X)
A__P₁(s₁(X)) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)
MARK₁(p₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__p₁(0) -> 0
a__p₁(s₁(X)) -> mark₁(X)
a__leq₂(0, Y) -> true
a__leq₂(s₁(X), 0) -> false
a__leq₂(s₁(X), s₁(Y)) -> a__leq₂(mark₁(X), mark₁(Y))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__diff₂(X, Y) -> a__if₃(a__leq₂(mark₁(X), mark₁(Y)), 0, s₁(diff₂(p₁(X), Y)))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(leq₂(X1, X2)) -> a__leq₂(mark₁(X1), mark₁(X2))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(diff₂(X1, X2)) -> a__diff₂(mark₁(X1), mark₁(X2))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__p₁(X) -> p₁(X)
a__leq₂(X1, X2) -> leq₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__diff₂(X1, X2) -> diff₂(X1, X2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

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

MARK₁(diff₂(X1, X2)) -> MARK₁(X2)
MARK₁(diff₂(X1, X2)) -> A__DIFF₂(mark₁(X1), mark₁(X2))
A__LEQ₂(s₁(X), s₁(Y)) -> MARK₁(X)
A__DIFF₂(X, Y) -> A__IF₃(a__leq₂(mark₁(X), mark₁(Y)), 0, s₁(diff₂(p₁(X), Y)))
A__LEQ₂(s₁(X), s₁(Y)) -> A__LEQ₂(mark₁(X), mark₁(Y))
MARK₁(diff₂(X1, X2)) -> MARK₁(X1)
A__IF₃(false, X, Y) -> MARK₁(Y)
A__DIFF₂(X, Y) -> A__LEQ₂(mark₁(X), mark₁(Y))
MARK₁(p₁(X)) -> A__P₁(mark₁(X))
MARK₁(leq₂(X1, X2)) -> A__LEQ₂(mark₁(X1), mark₁(X2))
A__DIFF₂(X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__DIFF₂(X, Y) -> MARK₁(Y)
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(leq₂(X1, X2)) -> MARK₁(X2)
MARK₁(leq₂(X1, X2)) -> MARK₁(X1)
A__LEQ₂(s₁(X), s₁(Y)) -> MARK₁(Y)
A__IF₃(true, X, Y) -> MARK₁(X)
A__P₁(s₁(X)) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)
MARK₁(p₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__p₁(0) -> 0
a__p₁(s₁(X)) -> mark₁(X)
a__leq₂(0, Y) -> true
a__leq₂(s₁(X), 0) -> false
a__leq₂(s₁(X), s₁(Y)) -> a__leq₂(mark₁(X), mark₁(Y))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__diff₂(X, Y) -> a__if₃(a__leq₂(mark₁(X), mark₁(Y)), 0, s₁(diff₂(p₁(X), Y)))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(leq₂(X1, X2)) -> a__leq₂(mark₁(X1), mark₁(X2))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(diff₂(X1, X2)) -> a__diff₂(mark₁(X1), mark₁(X2))
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(mark₁(X))
mark₁(true) -> true
mark₁(false) -> false
a__p₁(X) -> p₁(X)
a__leq₂(X1, X2) -> leq₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__diff₂(X1, X2) -> diff₂(X1, X2)

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