Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
2(x, x) -> e
/2(x, x) -> e
.2(e, x) -> x
.2(x, e) -> x
2(e, x) -> x
/2(x, e) -> x
.2(x, 2(x, y)) -> y
.2(/2(y, x), x) -> y
2(x, .2(x, y)) -> y
/2(.2(y, x), x) -> y
/2(x, 2(y, x)) -> y
2(/2(x, y), x) -> y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
2(x, x) -> e
/2(x, x) -> e
.2(e, x) -> x
.2(x, e) -> x
2(e, x) -> x
/2(x, e) -> x
.2(x, 2(x, y)) -> y
.2(/2(y, x), x) -> y
2(x, .2(x, y)) -> y
/2(.2(y, x), x) -> y
/2(x, 2(y, x)) -> y
2(/2(x, y), x) -> y
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
2(x, x) -> e
/2(x, x) -> e
.2(e, x) -> x
.2(x, e) -> x
2(e, x) -> x
/2(x, e) -> x
.2(x, 2(x, y)) -> y
.2(/2(y, x), x) -> y
2(x, .2(x, y)) -> y
/2(.2(y, x), x) -> y
/2(x, 2(y, x)) -> y
2(/2(x, y), x) -> y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
2(x, x) -> e
/2(x, x) -> e
.2(e, x) -> x
.2(x, e) -> x
2(e, x) -> x
/2(x, e) -> x
.2(x, 2(x, y)) -> y
.2(/2(y, x), x) -> y
2(x, .2(x, y)) -> y
/2(.2(y, x), x) -> y
/2(x, 2(y, x)) -> y
2(/2(x, y), x) -> y
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.