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:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
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:
F(x, c(y)) → F(y, y)
F(s(x), y) → F(x, s(c(y)))
F(x, c(y)) → F(x, s(f(y, y)))
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
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:
F(x, c(y)) → F(y, y)
F(s(x), y) → F(x, s(c(y)))
F(x, c(y)) → F(x, s(f(y, y)))
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
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
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(x, c(y)) → F(y, y)
F(s(x), y) → F(x, s(c(y)))
F(x, c(y)) → F(x, s(f(y, y)))
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y) → F(x, s(c(y)))
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
F(s(x), y) → F(x, s(c(y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2) = x1
s(x1) = s(x1)
Recursive Path Order [2].
Precedence:
trivial
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(x, c(y)) → F(y, y)
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(y)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
F(x, c(y)) → F(y, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2) = x2
c(x1) = c(x1)
Recursive Path Order [2].
Precedence:
trivial
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x, c(y)) → f(x, s(f(y, y)))
f(s(x), y) → f(x, s(c(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.