(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(0) → 0
+(x, 0) → x
+(0, y) → y
+(minus(1), 1) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)
+(minus(+(x, 1)), 1) → minus(x)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(+(x1, x2)) = x1 + x2   
POL(0) = 0   
POL(1) = 1   
POL(minus(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

+(minus(1), 1) → 0
+(minus(+(x, 1)), 1) → minus(x)


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(0) → 0
+(x, 0) → x
+(0, y) → y
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(+(x1, x2)) = x1 + x2   
POL(0) = 1   
POL(minus(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

+(x, 0) → x
+(0, y) → y


(4) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(0) → 0
minus(minus(x)) → x
+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(+(x1, x2)) = x1 + 2·x2   
POL(0) = 0   
POL(minus(x1)) = 2 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

minus(0) → 0
minus(minus(x)) → x


(6) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

+(x, minus(y)) → minus(+(minus(x), y))
+(x, +(y, z)) → +(+(x, y), z)

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(+(x1, x2)) = 1 + x1 + 2·x2   
POL(minus(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

+(x, +(y, z)) → +(+(x, y), z)


(8) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

+(x, minus(y)) → minus(+(minus(x), y))

Q is empty.

(9) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(10) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

+(x, minus(y)) → minus(+(minus(x), y))

The set Q consists of the following terms:

+(x0, minus(x1))

(11) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

+1(x, minus(y)) → +1(minus(x), y)

The TRS R consists of the following rules:

+(x, minus(y)) → minus(+(minus(x), y))

The set Q consists of the following terms:

+(x0, minus(x1))

We have to consider all minimal (P,Q,R)-chains.

(13) 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.

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

+1(x, minus(y)) → +1(minus(x), y)

R is empty.
The set Q consists of the following terms:

+(x0, minus(x1))

We have to consider all minimal (P,Q,R)-chains.

(15) 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].

+(x0, minus(x1))

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

+1(x, minus(y)) → +1(minus(x), y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(17) 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, minus(y)) → +1(minus(x), y)
    The graph contains the following edges 2 > 2

(18) TRUE