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__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
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:
MARK(if(X1, X2, X3)) → MARK(X1)
A__F(X) → A__IF(mark(X), c, f(true))
MARK(f(X)) → A__F(mark(X))
A__IF(false, X, Y) → MARK(Y)
A__F(X) → MARK(X)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X2)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), mark(X2), X3)
MARK(f(X)) → MARK(X)
The TRS R consists of the following rules:
a__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
MARK(if(X1, X2, X3)) → MARK(X1)
A__F(X) → A__IF(mark(X), c, f(true))
MARK(f(X)) → A__F(mark(X))
A__IF(false, X, Y) → MARK(Y)
A__F(X) → MARK(X)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X2)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), mark(X2), X3)
MARK(f(X)) → MARK(X)
The TRS R consists of the following rules:
a__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
MARK(if(X1, X2, X3)) → MARK(X1)
A__F(X) → A__IF(mark(X), c, f(true))
MARK(f(X)) → A__F(mark(X))
A__F(X) → MARK(X)
A__IF(false, X, Y) → MARK(Y)
A__IF(true, X, Y) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X2)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), mark(X2), X3)
MARK(f(X)) → MARK(X)
The TRS R consists of the following rules:
a__f(X) → a__if(mark(X), c, f(true))
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(f(X)) → a__f(mark(X))
mark(if(X1, X2, X3)) → a__if(mark(X1), mark(X2), X3)
mark(c) → c
mark(true) → true
mark(false) → false
a__f(X) → f(X)
a__if(X1, X2, X3) → if(X1, X2, X3)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.