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:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
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:
F(x, c(x), c(y)) → F(y, x, y)
F(s(x), y, z) → F(x, s(c(y)), c(z))
F(x, c(x), c(y)) → F(y, y, f(y, x, y))
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
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:
F(x, c(x), c(y)) → F(y, x, y)
F(s(x), y, z) → F(x, s(c(y)), c(z))
F(x, c(x), c(y)) → F(y, y, f(y, x, y))
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y, z) → F(x, s(c(y)), c(z))
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(s(x), y, z) → F(x, s(c(y)), c(z))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- F(s(x), y, z) → F(x, s(c(y)), c(z))
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(x, c(x), c(y)) → F(y, x, y)
F(x, c(x), c(y)) → F(y, y, f(y, x, y))
The TRS R consists of the following rules:
f(x, c(x), c(y)) → f(y, y, f(y, x, y))
f(s(x), y, z) → f(x, s(c(y)), c(z))
f(c(x), x, y) → c(y)
g(x, y) → x
g(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.