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
↳ 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)))
Q is empty.
Using Dependency Pairs [1,15] 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)))
Q is empty.
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:
+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)))
Q is empty.
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
↳ DependencyPairsProof
↳ 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)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
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: Polynomial interpretation [25,35]:
POL(s(x1)) = 3 + (3)x_1
POL(0) = 0
POL(+1(x1, x2)) = (3)x_2
POL(+(x1, x2)) = (2)x_1 + (2)x_2
The value of delta used in the strict ordering is 9.
The following usable rules [17] were oriented:
+(0, y) → y
+(s(x), 0) → s(x)
↳ QTRS
↳ DependencyPairsProof
↳ 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)))
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.