(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 sequence

The DP semiunifies directly so there is only one rewrite step from 1(x) to 1(x).



(18) NO