(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
+(1, x) → +(+(0, 1), x)
+(0, x) → x
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:
+(1, x) → +(+(0, 1), x)
+(0, x) → x
The set Q consists of the following terms:
+(1, x0)
+(0, x0)
(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(1, x) → +1(+(0, 1), x)
+1(1, x) → +1(0, 1)
The TRS R consists of the following rules:
+(1, x) → +(+(0, 1), x)
+(0, x) → x
The set Q consists of the following terms:
+(1, x0)
+(0, x0)
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(1, x) → +1(+(0, 1), x)
The TRS R consists of the following rules:
+(1, x) → +(+(0, 1), x)
+(0, x) → x
The set Q consists of the following terms:
+(1, x0)
+(0, x0)
We have to consider all minimal (P,Q,R)-chains.
(7) 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.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(1, x) → +1(+(0, 1), x)
The TRS R consists of the following rules:
+(0, x) → x
The set Q consists of the following terms:
+(1, x0)
+(0, x0)
We have to consider all minimal (P,Q,R)-chains.
(9) Rewriting (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
+1(
1,
x) →
+1(
+(
0,
1),
x) at position [0] we obtained the following new rules [LPAR04]:
+1(1, x) → +1(1, x)
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(1, x) → +1(1, x)
The TRS R consists of the following rules:
+(0, x) → x
The set Q consists of the following terms:
+(1, x0)
+(0, x0)
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(1, x) → +1(1, x)
R is empty.
The set Q consists of the following terms:
+(1, x0)
+(0, x0)
We have to consider all minimal (P,Q,R)-chains.
(13) 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].
+(1, x0)
+(0, x0)
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
+1(1, x) → +1(1, x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) ATransformationProof (EQUIVALENT transformation)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
1(x) → 1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.
s =
1(
x) evaluates to t =
1(
x)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from 1(x) to 1(x).
(18) FALSE