Termination w.r.t. Q proof of TRS/AG01#3.6b.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

Q restricted rewrite system:
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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(x0), s(x1))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF_GCD(false, s(x), s(y)) → GCD(minus(y, x), s(x))
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
IF_GCD(true, s(x), s(y)) → MINUS(x, y)
MINUS(s(x), y) → LE(s(x), y)
LE(s(x), s(y)) → LE(x, y)
GCD(s(x), s(y)) → LE(y, x)
IF_GCD(true, s(x), s(y)) → GCD(minus(x, y), s(y))
GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
IF_MINUS(false, s(x), y) → MINUS(x, y)
IF_GCD(false, s(x), s(y)) → MINUS(y, x)

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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(x0), s(x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

IF_GCD(false, s(x), s(y)) → GCD(minus(y, x), s(x))
MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
IF_GCD(true, s(x), s(y)) → MINUS(x, y)
MINUS(s(x), y) → LE(s(x), y)
LE(s(x), s(y)) → LE(x, y)
GCD(s(x), s(y)) → LE(y, x)
IF_GCD(true, s(x), s(y)) → GCD(minus(x, y), s(y))
GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
IF_MINUS(false, s(x), y) → MINUS(x, y)
IF_GCD(false, s(x), s(y)) → MINUS(y, x)

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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
IF_GCD(false, s(x), s(y)) → GCD(minus(y, x), s(x))
IF_GCD(true, s(x), s(y)) → MINUS(x, y)
MINUS(s(x), y) → LE(s(x), y)
LE(s(x), s(y)) → LE(x, y)
GCD(s(x), s(y)) → LE(y, x)
IF_GCD(true, s(x), s(y)) → GCD(minus(x, y), s(y))
IF_MINUS(false, s(x), y) → MINUS(x, y)
GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
IF_GCD(false, s(x), s(y)) → MINUS(y, x)

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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = LE(x1)
s(x1) = s(x1)

Lexicographic path order with status [19].
Precedence:

s₁ > LE₁

Status:

LE₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

Q DP problem:
P is empty.
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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, 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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
IF_MINUS(false, s(x), y) → MINUS(x, y)

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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

IF_MINUS(false, s(x), y) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)

Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = x1
s(x1) = s(x1)
IF_MINUS(x1, x2, x3) = x2
le(x1, x2) = le(x1, x2)
false = false
0 = 0
true = true

Lexicographic path order with status [19].
Precedence:

s₁ > le₂
false > le₂
0 > le₂
true > le₂

Status:

true: multiset
false: multiset
s₁: multiset
0: multiset
le₂: [2,1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                  ↳ QDP

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

MINUS(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)

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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

IF_GCD(false, s(x), s(y)) → GCD(minus(y, x), s(x))
IF_GCD(true, s(x), s(y)) → GCD(minus(x, y), s(y))
GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))

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)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(0, x0)
minus(s(x0), x1)
if_minus(true, s(x0), x1)
if_minus(false, s(x0), x1)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if_gcd(true, s(x0), s(x1))
if_gcd(false, s(x0), s(x1))

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