(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) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(a, +(b, z)) → +1(a, z)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) ATransformationProof (EQUIVALENT transformation)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
a(b(z)) → a(z)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- a(b(z)) → a(z)
The graph contains the following edges 1 > 1
(11) TRUE
(12) 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.
(13) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(14) 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))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- +1(+(x, y), z) → +1(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
- +1(+(x, y), z) → +1(x, +(y, z))
The graph contains the following edges 1 > 1
(16) TRUE
(17) 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.
(18) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(19) 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)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- F(+(x, y), z) → F(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
- F(+(x, y), z) → F(x, z)
The graph contains the following edges 1 > 1, 2 >= 2
(21) TRUE