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:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y
The set Q consists of the following terms:
f(0, s(0), x0)
+(x0, s(x1))
+(x0, 0)
g(x0, x1)
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y
The set Q consists of the following terms:
f(0, s(0), x0)
+(x0, s(x1))
+(x0, 0)
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, s(0), X) → F(X, +(X, X), X)
+1(X, s(Y)) → +1(X, Y)
F(0, s(0), X) → +1(X, X)
The TRS R consists of the following rules:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y
The set Q consists of the following terms:
f(0, s(0), x0)
+(x0, s(x1))
+(x0, 0)
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(0, s(0), X) → F(X, +(X, X), X)
+1(X, s(Y)) → +1(X, Y)
F(0, s(0), X) → +1(X, X)
The TRS R consists of the following rules:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y
The set Q consists of the following terms:
f(0, s(0), x0)
+(x0, s(x1))
+(x0, 0)
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:
+1(X, s(Y)) → +1(X, Y)
The TRS R consists of the following rules:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y
The set Q consists of the following terms:
f(0, s(0), x0)
+(x0, s(x1))
+(x0, 0)
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:
+1(X, s(Y)) → +1(X, Y)
R is empty.
The set Q consists of the following terms:
f(0, s(0), x0)
+(x0, s(x1))
+(x0, 0)
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, s(0), x0)
+(x0, s(x1))
+(x0, 0)
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:
+1(X, s(Y)) → +1(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:
- +1(X, s(Y)) → +1(X, Y)
The graph contains the following edges 1 >= 1, 2 > 2
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(0, s(0), X) → F(X, +(X, X), X)
The TRS R consists of the following rules:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y
The set Q consists of the following terms:
f(0, s(0), x0)
+(x0, s(x1))
+(x0, 0)
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(0, s(0), X) → F(X, +(X, X), X)
The TRS R consists of the following rules:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
The set Q consists of the following terms:
f(0, s(0), x0)
+(x0, s(x1))
+(x0, 0)
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, s(0), x0)
g(x0, x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
F(0, s(0), X) → F(X, +(X, X), X)
The TRS R consists of the following rules:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
The set Q consists of the following terms:
+(x0, s(x1))
+(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule F(0, s(0), X) → F(X, +(X, X), X) at position [1] we obtained the following new rules:
F(0, s(0), 0) → F(0, 0, 0)
F(0, s(0), s(x1)) → F(s(x1), s(+(s(x1), x1)), s(x1))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(0, s(0), 0) → F(0, 0, 0)
F(0, s(0), s(x1)) → F(s(x1), s(+(s(x1), x1)), s(x1))
The TRS R consists of the following rules:
+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
The set Q consists of the following terms:
+(x0, s(x1))
+(x0, 0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.