(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(s(x), y) → +1(x, y)
+1(p(x), y) → +1(x, y)
MINUS(s(x)) → MINUS(x)
MINUS(p(x)) → MINUS(x)
*1(s(x), y) → +1(*(x, y), y)
*1(s(x), y) → *1(x, y)
*1(p(x), y) → +1(*(x, y), minus(y))
*1(p(x), y) → *1(x, y)
*1(p(x), y) → MINUS(y)
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)
R is empty.
The set Q consists of the following terms:
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(10) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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:
- MINUS(p(x)) → MINUS(x)
The graph contains the following edges 1 > 1
- MINUS(s(x)) → MINUS(x)
The graph contains the following edges 1 > 1
(13) TRUE
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(p(x), y) → +1(x, y)
+1(s(x), y) → +1(x, y)
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(p(x), y) → +1(x, y)
+1(s(x), y) → +1(x, y)
R is empty.
The set Q consists of the following terms:
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(17) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(p(x), y) → +1(x, y)
+1(s(x), y) → +1(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) 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(p(x), y) → +1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
- +1(s(x), y) → +1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
(20) TRUE
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(p(x), y) → *1(x, y)
*1(s(x), y) → *1(x, y)
The TRS R consists of the following rules:
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))
The set Q consists of the following terms:
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(22) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(p(x), y) → *1(x, y)
*1(s(x), y) → *1(x, y)
R is empty.
The set Q consists of the following terms:
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(24) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
*1(p(x), y) → *1(x, y)
*1(s(x), y) → *1(x, y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) 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(p(x), y) → *1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
- *1(s(x), y) → *1(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
(27) TRUE