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))
+(0, s(y)) → s(y)
s(+(0, y)) → s(y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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:
+1(x, s(y)) → +1(x, y)
+1(x, s(y)) → S(+(x, y))
S(+(0, y)) → S(y)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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:
+1(x, s(y)) → +1(x, y)
+1(x, s(y)) → S(+(x, y))
S(+(0, y)) → S(y)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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:
+1(x, s(y)) → +1(x, y)
+1(x, s(y)) → S(+(x, y))
S(+(0, y)) → S(y)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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:
S(+(0, y)) → S(y)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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.
S(+(0, y)) → S(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
+(x1, x2) = +(x1, x2)
0 = 0
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:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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:
+1(x, s(y)) → +1(x, y)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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.
+1(x, s(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+1(x1, x2) = x2
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
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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.