(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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)) → S(+(x, y))
+1(x, s(y)) → +1(x, y)
S(+(0, y)) → S(y)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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:
S(+(0, y)) → S(y)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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.
S(+(0, y)) → S(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(
x1) =
S(
x1)
+(
x1,
x2) =
+(
x1,
x2)
0 =
0
s(
x1) =
s
Lexicographic path order with status [LPO].
Precedence:
trivial
Status:
s: []
0: []
+2: [1,2]
S1: [1]
The following usable rules [FROCOS05] were oriented:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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))
+(0, s(y)) → s(y)
s(+(0, y)) → s(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:
+1(x, s(y)) → +1(x, y)
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(y)
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, 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) =
+1(
x2)
s(
x1) =
s(
x1)
+(
x1,
x2) =
+(
x1,
x2)
0 =
0
Lexicographic path order with status [LPO].
Precedence:
+^11 > s1
+2 > s1
0 > s1
Status:
s1: [1]
0: []
+2: [1,2]
+^11: [1]
The following usable rules [FROCOS05] were oriented:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(y)
(12) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, s(y)) → s(y)
s(+(0, y)) → s(y)
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