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:
g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), x))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G(g(x)) → G(h(g(x)))
H(h(x)) → H(f(h(x), x))
G(g(x)) → H(g(x))
The TRS R consists of the following rules:
g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), 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:
G(g(x)) → G(h(g(x)))
H(h(x)) → H(f(h(x), x))
G(g(x)) → H(g(x))
The TRS R consists of the following rules:
g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
G(g(x)) → G(h(g(x)))
The TRS R consists of the following rules:
g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
G(g(x)) → G(h(g(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( f(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
g(g(x)) → g(h(g(x)))
g(h(g(x))) → g(x)
h(h(x)) → h(f(h(x), x))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
g(h(g(x))) → g(x)
g(g(x)) → g(h(g(x)))
h(h(x)) → h(f(h(x), 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.