Termination w.r.t. Q proof of /tmp/tmpoKgW7o/wiehe01.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:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(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:

PLUS(id(x), s(y)) → GT(s(y), y)
PLUS(s(x), x) → IF(gt(x, x), id(x), id(x))
GT(s(x), s(y)) → GT(x, y)
PLUS(s(x), s(y)) → IF(not(gt(x, y)), id(x), id(y))
PLUS(s(x), s(y)) → NOT(gt(x, y))
PLUS(s(x), x) → GT(x, x)
PLUS(s(x), s(y)) → ID(y)
PLUS(s(x), x) → ID(x)
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(s(x), s(y)) → GT(x, y)
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, s(y)) → PLUS(times(x, y), x)
PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
NOT(x) → IF(x, false, true)
PLUS(s(x), s(y)) → ID(x)
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
PLUS(id(x), s(y)) → IF(gt(s(y), y), y, s(y))
PLUS(s(x), s(y)) → IF(gt(x, y), x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(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:

PLUS(id(x), s(y)) → GT(s(y), y)
PLUS(s(x), x) → IF(gt(x, x), id(x), id(x))
GT(s(x), s(y)) → GT(x, y)
PLUS(s(x), s(y)) → IF(not(gt(x, y)), id(x), id(y))
PLUS(s(x), s(y)) → NOT(gt(x, y))
PLUS(s(x), x) → GT(x, x)
PLUS(s(x), s(y)) → ID(y)
PLUS(s(x), x) → ID(x)
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))
PLUS(s(x), s(y)) → GT(x, y)
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, s(y)) → PLUS(times(x, y), x)
PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
NOT(x) → IF(x, false, true)
PLUS(s(x), s(y)) → ID(x)
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
PLUS(id(x), s(y)) → IF(gt(s(y), y), y, s(y))
PLUS(s(x), s(y)) → IF(gt(x, y), x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(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 13 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(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:

GT(s(x), s(y)) → GT(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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))
PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
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.

PLUS(s(x), s(y)) → PLUS(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))

The remaining pairs can at least be oriented weakly.

PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(PLUS(x₁, x₂)) = max(x₁, x₂)
POL(false) = 0
POL(gt(x₁, x₂)) = 0
POL(id(x₁)) = x₁
POL(if(x₁, x₂, x₃)) = max(x₂, x₃)
POL(not(x₁)) = 0
POL(s(x₁)) = 1 + x₁
POL(true) = 0
POL(zero) = 0

The following usable rules [17] were oriented:

if(false, x, y) → y
id(x) → x
if(true, x, y) → x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

Q is empty.
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.

PLUS(s(x), x) → PLUS(if(gt(x, x), id(x), id(x)), s(x))

The remaining pairs can at least be oriented weakly.

PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))

Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = PLUS(x1)
id(x1) = x1
s(x1) = s(x1)
if(x1, x2, x3) = if(x1, x2, x3)
gt(x1, x2) = gt(x1, x2)
false = false
true = true
zero = zero

Recursive path order with status [2].
Quasi-Precedence:

s₁ > if₃ > [PLUS₁, gt₂]
[true, zero] > false > [PLUS₁, gt₂]

Status:

if₃: [1,3,2]
zero: multiset
true: multiset
false: multiset
s₁: [1]
gt₂: [2,1]
PLUS₁: multiset

The following usable rules [17] were oriented:

if(false, x, y) → y
id(x) → x
if(true, x, y) → x
gt(s(x), s(y)) → gt(x, y)
gt(s(x), zero) → true
gt(zero, y) → false

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

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:

PLUS(id(x), s(y)) → PLUS(x, if(gt(s(y), y), y, s(y)))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

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

TIMES(x, s(y)) → TIMES(x, y)

The TRS R consists of the following rules:

times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(s(x), s(y)) → s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))
plus(s(x), x) → plus(if(gt(x, x), id(x), id(x)), s(x))
plus(zero, y) → y
plus(id(x), s(y)) → s(plus(x, if(gt(s(y), y), y, s(y))))
id(x) → x
if(true, x, y) → x
if(false, x, y) → y
not(x) → if(x, false, true)
gt(s(x), zero) → true
gt(zero, y) → false
gt(s(x), s(y)) → gt(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

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

TIMES(x, s(y)) → TIMES(x, y)

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

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