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(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
The set Q consists of the following terms:
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
↳ 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
The set Q consists of the following terms:
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
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
The set Q consists of the following terms:
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
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
The set Q consists of the following terms:
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.
↳ 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
The set Q consists of the following terms:
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ 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.
The set Q consists of the following terms:
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ 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
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(x, c(x), c(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
The set Q consists of the following terms:
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(x, c(x), c(y)) → F(y, x, y)
R is empty.
The set Q consists of the following terms:
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
f(x0, c(x0), c(x1))
f(c(x0), x0, x1)
f(s(x0), x1, x2)
g(x0, x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
F(x, c(x), c(y)) → F(y, x, y)
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(x, c(x), c(y)) → F(y, x, y)
The graph contains the following edges 3 > 1, 1 >= 2, 2 > 2, 3 > 3