Termination w.r.t. Q proof of /tmp/tmpBiOY6L/perfect2.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:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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

F(s(x), s(y), z, u) → MINUS(y, x)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), s(y), z, u) → IF(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
MINUS(s(x), s(y)) → MINUS(x, y)
F(s(x), s(y), z, u) → LE(x, y)
F(s(x), 0, z, u) → MINUS(z, s(x))
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F(s(x), s(y), z, u) → MINUS(y, x)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), s(y), z, u) → IF(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
MINUS(s(x), s(y)) → MINUS(x, y)
F(s(x), s(y), z, u) → LE(x, y)
F(s(x), 0, z, u) → MINUS(z, s(x))
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

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

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), 0) → s(x)
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(x)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(x)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)

The remaining pairs can at least be oriented weakly.

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(F(x₁, x₂, x₃, x₄)) = x₁
POL(minus(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                            ↳ QDPOrderProof

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

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(x)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(F(x₁, x₂, x₃, x₄)) = x₂
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(x)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

                            ↳ QDPOrderProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(x)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( minus(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

/	0	0	\
\	0	0	/

x₂

M( s(x₁) ) =

/	0	\
\	1	/

/	0	0	\
\	0	1	/

x₁

M( 0 ) =

/	0	\
\	0	/

Tuple symbols:

M( F(x₁, ..., x₄) ) =

[

]

x₁

[

]

x₂

[

]

x₃

[

]

x₄

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(x)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                            ↳ QDPOrderProof

                              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(s(x), 0) → s(x)

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), 0)
minus(s(x0), s(x1))

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