(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(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(.(x, y), z) → ++1(y, z)
++1(++(x, y), z) → ++1(x, ++(y, z))
++1(++(x, y), z) → ++1(y, z)
The TRS R consists of the following rules:
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) 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(x, ++(y, 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) =
x1
.(
x1,
x2) =
x2
++(
x1,
x2) =
++(
x1,
x2)
nil =
nil
Recursive path order with status [RPO].
Precedence:
trivial
Status:
++2: [1,2]
The following usable rules [FROCOS05] were oriented:
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
++1(.(x, y), z) → ++1(y, z)
The TRS R consists of the following rules:
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))
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)
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) =
.(
x2)
++(
x1,
x2) =
++(
x1,
x2)
nil =
nil
Recursive path order with status [RPO].
Precedence:
++2 > .1
Status:
++2: [1,2]
The following usable rules [FROCOS05] were oriented:
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(++(x, y), z) → ++(x, ++(y, z))
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