Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

-1(x, s(y)) → P(s(y))
-1(x, s(y)) → -1(x, p(s(y)))

The TRS R consists of the following rules:

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

-1(x, s(y)) → P(s(y))
-1(x, s(y)) → -1(x, p(s(y)))

The TRS R consists of the following rules:

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

-1(x, s(y)) → -1(x, p(s(y)))

The TRS R consists of the following rules:

-(0, y) → 0
-(x, 0) → x
-(x, s(y)) → if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) → 0
p(s(x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ MNOCProof

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

-1(x, s(y)) → -1(x, p(s(y)))

The TRS R consists of the following rules:

p(s(x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ MNOCProof
                ↳ QDP
                  ↳ RuleRemovalProof

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

-1(x, s(y)) → -1(x, p(s(y)))

The TRS R consists of the following rules:

p(s(x)) → x

The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

p(s(x)) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(-1(x1, x2)) = x1 + x2   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ MNOCProof
                ↳ QDP
                  ↳ RuleRemovalProof
                    ↳ QDP
                      ↳ DependencyGraphProof

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

-1(x, s(y)) → -1(x, p(s(y)))

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

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.