(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))
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:
F(s(x)) → G(x, s(x))
G(s(x), y) → G(x, +(y, s(x)))
G(s(x), y) → +1(y, s(x))
+1(x, s(y)) → +1(x, y)
G(s(x), y) → G(x, s(+(y, x)))
G(s(x), y) → +1(y, x)
The TRS R consists of the following rules:
f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))
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 3 less nodes.
(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:
f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPSizeChangeProof (EQUIVALENT transformation)
We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.
Order:Homeomorphic Embedding Order
AFS:
s(x1) = s(x1)
From the DPs we obtained the following set of size-change graphs:
- +1(x, s(y)) → +1(x, y) (allowed arguments on rhs = {1, 2})
The graph contains the following edges 1 >= 1, 2 > 2
We oriented the following set of usable rules [AAECC05,FROCOS05].
none
(7) TRUE
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
G(s(x), y) → G(x, s(+(y, x)))
G(s(x), y) → G(x, +(y, s(x)))
The TRS R consists of the following rules:
f(0) → 1
f(s(x)) → g(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) QDPSizeChangeProof (EQUIVALENT transformation)
We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.
Order:Homeomorphic Embedding Order
AFS:
s(x1) = s(x1)
From the DPs we obtained the following set of size-change graphs:
- G(s(x), y) → G(x, s(+(y, x))) (allowed arguments on rhs = {1})
The graph contains the following edges 1 > 1
- G(s(x), y) → G(x, +(y, s(x))) (allowed arguments on rhs = {1})
The graph contains the following edges 1 > 1
We oriented the following set of usable rules [AAECC05,FROCOS05].
none
(10) TRUE