Termination w.r.t. Q proof of /tmp/tmpbsO4vw/aprove4.trs

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:

p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

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:

IF(false, x, y) → P(x)
P(p(s(x))) → P(x)
MINUS(x, y) → LE(x, y)
IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), x, y)
LE(p(s(x)), x) → LE(x, x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

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:

IF(false, x, y) → P(x)
P(p(s(x))) → P(x)
MINUS(x, y) → LE(x, y)
IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), x, y)
LE(p(s(x)), x) → LE(x, x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

LE(p(s(x)), x) → LE(x, x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

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

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

LE(p(s(x)), x) → LE(x, x)
LE(s(x), s(y)) → LE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

LE(s(x), s(y)) → LE(x, y)
The graph contains the following edges 1 > 1, 2 > 2
LE(p(s(x)), x) → LE(x, x)
The graph contains the following edges 1 > 1, 2 >= 1, 1 > 2, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

P(p(s(x))) → P(x)

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

P(p(s(x))) → P(x)

From the DPs we obtained the following set of size-change graphs:

P(p(s(x))) → P(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

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

IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), x, y)

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

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

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

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

IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), x, y)

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), x, y)

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), x, y)

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MINUS(x, y) → IF(le(x, y), x, y) at position [0] we obtained the following new rules:

MINUS(0, x0) → IF(true, 0, x0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
MINUS(s(x0), 0) → IF(false, s(x0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, x, y) → MINUS(p(x), y)
MINUS(0, x0) → IF(true, 0, x0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, x, y) → MINUS(p(x), y)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF(false, x, y) → MINUS(p(x), y) at position [0] we obtained the following new rules:

IF(false, 0, y1) → MINUS(s(s(0)), y1)
IF(false, s(x0), y1) → MINUS(x0, y1)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

IF(false, 0, y1) → MINUS(s(s(0)), y1)
MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(x0), y1) → MINUS(x0, y1)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(x0), y1) → MINUS(x0, y1)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

p(0) → s(s(0))
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(x0), y1) → MINUS(x0, y1)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

p(0)
p(s(x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(x0), y1) → MINUS(x0, y1)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF(false, s(x0), y1) → MINUS(x0, y1) we obtained the following new rules:

IF(false, s(z0), s(z1)) → MINUS(z0, s(z1))
IF(false, s(z0), 0) → MINUS(z0, 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
IF(false, s(z0), s(z1)) → MINUS(z0, s(z1))
IF(false, s(z0), 0) → MINUS(z0, 0)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ AND

                                                        ↳ QDP

                                                          ↳ ForwardInstantiation

                                                        ↳ QDP

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

MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
IF(false, s(z0), s(z1)) → MINUS(z0, s(z1))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF(false, s(z0), s(z1)) → MINUS(z0, s(z1)) we obtained the following new rules:

IF(false, s(s(y_0)), s(x1)) → MINUS(s(y_0), s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ AND

                                                        ↳ QDP

                                                          ↳ ForwardInstantiation

                                                            ↳ QDP

                                                              ↳ ForwardInstantiation

                                                        ↳ QDP

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

IF(false, s(s(y_0)), s(x1)) → MINUS(s(y_0), s(x1))
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1)) we obtained the following new rules:

MINUS(s(s(y_1)), s(x1)) → IF(le(s(y_1), x1), s(s(y_1)), s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ AND

                                                        ↳ QDP

                                                          ↳ ForwardInstantiation

                                                            ↳ QDP

                                                              ↳ ForwardInstantiation

                                                                ↳ QDP

                                                                  ↳ QDPSizeChangeProof

                                                        ↳ QDP

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

MINUS(s(s(y_1)), s(x1)) → IF(le(s(y_1), x1), s(s(y_1)), s(x1))
IF(false, s(s(y_0)), s(x1)) → MINUS(s(y_0), s(x1))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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(s(s(y_1)), s(x1)) → IF(le(s(y_1), x1), s(s(y_1)), s(x1))
The graph contains the following edges 1 >= 2, 2 >= 3
IF(false, s(s(y_0)), s(x1)) → MINUS(s(y_0), s(x1))
The graph contains the following edges 2 > 1, 3 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ AND

                                                        ↳ QDP

                                                        ↳ QDP

                                                          ↳ UsableRulesProof

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(z0), 0) → MINUS(z0, 0)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ AND

                                                        ↳ QDP

                                                        ↳ QDP

                                                          ↳ UsableRulesProof

                                                            ↳ QDP

                                                              ↳ QReductionProof

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(z0), 0) → MINUS(z0, 0)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ AND

                                                        ↳ QDP

                                                        ↳ QDP

                                                          ↳ UsableRulesProof

                                                            ↳ QDP

                                                              ↳ QReductionProof

                                                                ↳ QDP

                                                                  ↳ ForwardInstantiation

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(z0), 0) → MINUS(z0, 0)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF(false, s(z0), 0) → MINUS(z0, 0) we obtained the following new rules:

IF(false, s(s(y_0)), 0) → MINUS(s(y_0), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ AND

                                                        ↳ QDP

                                                        ↳ QDP

                                                          ↳ UsableRulesProof

                                                            ↳ QDP

                                                              ↳ QReductionProof

                                                                ↳ QDP

                                                                  ↳ ForwardInstantiation

                                                                    ↳ QDP

                                                                      ↳ ForwardInstantiation

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

MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(s(y_0)), 0) → MINUS(s(y_0), 0)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule MINUS(s(x0), 0) → IF(false, s(x0), 0) we obtained the following new rules:

MINUS(s(s(y_0)), 0) → IF(false, s(s(y_0)), 0)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ MNOCProof

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ UsableRulesProof

                                          ↳ QDP

                                            ↳ QReductionProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

                                                      ↳ AND

                                                        ↳ QDP

                                                        ↳ QDP

                                                          ↳ UsableRulesProof

                                                            ↳ QDP

                                                              ↳ QReductionProof

                                                                ↳ QDP

                                                                  ↳ ForwardInstantiation

                                                                    ↳ QDP

                                                                      ↳ ForwardInstantiation

                                                                        ↳ QDP

                                                                          ↳ QDPSizeChangeProof

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

IF(false, s(s(y_0)), 0) → MINUS(s(y_0), 0)
MINUS(s(s(y_0)), 0) → IF(false, s(s(y_0)), 0)

From the DPs we obtained the following set of size-change graphs:

MINUS(s(s(y_0)), 0) → IF(false, s(s(y_0)), 0)
The graph contains the following edges 1 >= 2, 2 >= 3
IF(false, s(s(y_0)), 0) → MINUS(s(y_0), 0)
The graph contains the following edges 2 > 1, 3 >= 2