(0) Obligation:
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
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(X, s(Y)) → +1(X, Y)
F(0, s(0), X) → F(X, +(X, X), X)
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
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
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
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
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.
Used ordering: Combined order from the following AFS and order.
+1(
x1,
x2) =
x2
s(
x1) =
s(
x1)
+(
x1,
x2) =
+(
x1,
x2)
0 =
0
f(
x1,
x2,
x3) =
f
g(
x1,
x2) =
g(
x1,
x2)
Lexicographic path order with status [LPO].
Quasi-Precedence:
[+2, 0, f] > s1
Status:
s1: [1]
+2: [1,2]
0: []
f: []
g2: [1,2]
The following usable rules [FROCOS05] were oriented:
+(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
(7) Obligation:
Q DP problem:
P is empty.
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
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(9) TRUE
(10) Obligation:
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
Q is empty.
We have to consider all minimal (P,Q,R)-chains.