(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))
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, y), z) → :1(x, :(y, z))
:1(:(x, y), z) → :1(y, z)
:1(+(x, y), z) → :1(x, z)
:1(+(x, y), z) → :1(y, z)
:1(z, +(x, f(y))) → :1(g(z, y), +(x, a))
The TRS R consists of the following rules:
:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))
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 1 SCC with 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
:1(:(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(x, :(y, z))
:1(+(x, y), z) → :1(x, z)
:1(+(x, y), z) → :1(y, z)
The TRS R consists of the following rules:
:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
:1(:(x, y), z) → :1(y, z)
:1(:(x, y), z) → :1(x, :(y, z))
:1(+(x, y), z) → :1(x, z)
:1(+(x, y), z) → :1(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
:1(
x1,
x2) =
:1(
x1)
:(
x1,
x2) =
:(
x1,
x2)
+(
x1,
x2) =
+(
x1,
x2)
f(
x1) =
f
g(
x1,
x2) =
g(
x1,
x2)
a =
a
Lexicographic path order with status [LPO].
Quasi-Precedence:
:^11 > :2 > [+2, f, g2]
:^11 > :2 > a
Status:
a: []
:2: [1,2]
f: []
g2: [2,1]
:^11: [1]
+2: [1,2]
The following usable rules [FROCOS05] were oriented:
none
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
:(:(x, y), z) → :(x, :(y, z))
:(+(x, y), z) → +(:(x, z), :(y, z))
:(z, +(x, f(y))) → :(g(z, y), +(x, a))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(8) TRUE