(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))
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(a, b) → +1(b, a)
+1(a, +(b, z)) → +1(b, +(a, z))
+1(a, +(b, z)) → +1(a, z)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)
F(+(x, y), z) → +1(f(x, z), f(y, z))
F(+(x, y), z) → F(x, z)
F(+(x, y), z) → F(y, z)
The TRS R consists of the following rules:
+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))
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 3 SCCs with 3 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(a, +(b, z)) → +1(a, z)
The TRS R consists of the following rules:
+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))
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(a, +(b, z)) → +1(a, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(
x1,
x2) =
x2
a =
a
+(
x1,
x2) =
+(
x1,
x2)
b =
b
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))
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:
+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
The TRS R consists of the following rules:
+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(
x1,
x2) =
x1
+(
x1,
x2) =
+(
x1,
x2)
a =
a
b =
b
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(12) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(14) TRUE
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(+(x, y), z) → F(y, z)
F(+(x, y), z) → F(x, z)
The TRS R consists of the following rules:
+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F(+(x, y), z) → F(y, z)
F(+(x, y), z) → F(x, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(
x1,
x2) =
x1
+(
x1,
x2) =
+(
x1,
x2)
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(17) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+(a, b) → +(b, a)
+(a, +(b, z)) → +(b, +(a, z))
+(+(x, y), z) → +(x, +(y, z))
f(a, y) → a
f(b, y) → b
f(+(x, y), z) → +(f(x, z), f(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(19) TRUE