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:
+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))
Q is empty.
The TRS is overlay and locally confluent. By [15] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), 0)
+(s(x0), s(x1))
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(s(x), s(y)) → +1(y, 0)
+1(s(x), s(y)) → +1(s(x), +(y, 0))
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), 0)
+(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
+1(s(x), s(y)) → +1(y, 0)
+1(s(x), s(y)) → +1(s(x), +(y, 0))
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), 0)
+(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
+1(s(x), s(y)) → +1(y, 0)
+1(s(x), s(y)) → +1(s(x), +(y, 0))
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), 0)
+(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
+1(s(x), s(y)) → +1(s(x), +(y, 0))
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), 0)
+(s(x0), s(x1))
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(s(x), s(y)) → +1(s(x), +(y, 0))
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)
+(x1, x2) = +(x1, x2)
0 = 0
Recursive path order with status [2].
Quasi-Precedence:
s1 > +2
s1 > 0
Status: trivial
The following usable rules [14] were oriented:
+(0, y) → y
+(s(x), 0) → s(x)
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), 0) → s(x)
+(s(x), s(y)) → s(+(s(x), +(y, 0)))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), 0)
+(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.