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(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is
h(0) → 0
h(g(x, y)) → y
The TRS R 2 is
f(0, 1, x) → f(h(x), h(x), x)
The signature Sigma is {f}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(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(0, 1, x) → H(x)
F(0, 1, x) → F(h(x), h(x), x)
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → H(x)
F(0, 1, x) → F(h(x), h(x), x)
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(h(x), h(x), x)
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(h(x), h(x), x)
The TRS R consists of the following rules:
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(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(0, 1, x0)
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(h(x), h(x), x)
The TRS R consists of the following rules:
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(0, 1, x) → F(h(x), h(x), x) at position [0] we obtained the following new rules:
F(0, 1, 0) → F(0, h(0), 0)
F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1))
F(0, 1, 0) → F(0, h(0), 0)
The TRS R consists of the following rules:
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1))
The TRS R consists of the following rules:
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1))
The TRS R consists of the following rules:
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1)) at position [1] we obtained the following new rules:
F(0, 1, g(x0, x1)) → F(x1, x1, g(x0, x1))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ Rewriting
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x0, x1)) → F(x1, x1, g(x0, x1))
The TRS R consists of the following rules:
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.