Termination w.r.t. Q proof of /tmp/tmpGJzwqP/kabasci05.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:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(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:

GCD(s(x), s(y)) → MINUS(max(x, y), min(x, transform(y)))
MIN(s(x), s(y)) → MIN(x, y)
TRANSFORM(cons(x, y)) → CONS(x, x)
GCD(s(x), s(y)) → MIN(x, transform(y))
TRANSFORM(s(x)) → TRANSFORM(x)
GCD(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → MIN(x, y)
TRANSFORM(cons(x, y)) → CONS(cons(x, x), x)
CONS(x, cons(y, s(z))) → CONS(y, x)
CONS(cons(x, z), s(y)) → TRANSFORM(x)
MINUS(s(x), s(y)) → MINUS(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
GCD(s(x), s(y)) → TRANSFORM(y)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(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:

GCD(s(x), s(y)) → MINUS(max(x, y), min(x, transform(y)))
MIN(s(x), s(y)) → MIN(x, y)
TRANSFORM(cons(x, y)) → CONS(x, x)
GCD(s(x), s(y)) → MIN(x, transform(y))
TRANSFORM(s(x)) → TRANSFORM(x)
GCD(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → MIN(x, y)
TRANSFORM(cons(x, y)) → CONS(cons(x, x), x)
CONS(x, cons(y, s(z))) → CONS(y, x)
CONS(cons(x, z), s(y)) → TRANSFORM(x)
MINUS(s(x), s(y)) → MINUS(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
GCD(s(x), s(y)) → TRANSFORM(y)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TRANSFORM(cons(x, y)) → CONS(cons(x, x), x)
CONS(x, cons(y, s(z))) → CONS(y, x)
CONS(cons(x, z), s(y)) → TRANSFORM(x)
TRANSFORM(cons(x, y)) → CONS(x, x)
TRANSFORM(s(x)) → TRANSFORM(x)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)

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.

TRANSFORM(cons(x, y)) → CONS(cons(x, x), x)
CONS(x, cons(y, s(z))) → CONS(y, x)
CONS(cons(x, z), s(y)) → TRANSFORM(x)
TRANSFORM(cons(x, y)) → CONS(x, x)
TRANSFORM(s(x)) → TRANSFORM(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(TRANSFORM(x₁)) = 1/4 + (9/4)x_1
POL(cons(x₁, x₂)) = 1/4 + (2)x_1 + x_2
POL(CONS(x₁, x₂)) = (5/4)x_1 + (3/4)x_2
POL(transform(x₁)) = 1/2 + (7/2)x_1
POL(s(x₁)) = 1/4 + x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

transform(x) → s(s(x))
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
transform(cons(x, y)) → cons(cons(x, x), x)
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)
transform(cons(x, y)) → y

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ 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:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(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

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

MAX(s(x), s(y)) → MAX(x, y)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

MAX(s(x), s(y)) → MAX(x, y)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

MIN(s(x), s(y)) → MIN(x, y)

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

MIN(s(x), s(y)) → MIN(x, y)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, transform(y))), s(min(x, y)))

The TRS R consists of the following rules:

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)

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.

GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, transform(y))), s(min(x, y)))

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

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

/	0	0	\
\	1	1	/

x₂

M( minus(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	0	0	\
\	0	0	/

x₂

M( max(x₁, x₂) ) =

/	0	\
\	1	/

/	1	0	\
\	0	1	/

x₁

/	1	0	\
\	0	1	/

x₂

M( transform(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( s(x₁) ) =

/	1	\
\	1	/

/	1	1	\
\	1	1	/

x₁

M( min(x₁, x₂) ) =

/	0	\
\	0	/

/	0	1	\
\	1	0	/

x₁

/	0	0	\
\	0	0	/

x₂

M( 0 ) =

/	0	\
\	0	/

Tuple symbols:

M( GCD(x₁, x₂) ) =

[

]

x₁

[

]

x₂

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:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

min(x, 0) → 0
min(0, y) → 0
min(s(x), s(y)) → s(min(x, y))
max(x, 0) → x
max(0, y) → y
max(s(x), s(y)) → s(max(x, y))
minus(x, 0) → x
minus(s(x), s(y)) → s(minus(x, y))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, transform(y))), s(min(x, y)))
transform(x) → s(s(x))
transform(cons(x, y)) → cons(cons(x, x), x)
transform(cons(x, y)) → y
transform(s(x)) → s(s(transform(x)))
cons(x, y) → y
cons(x, cons(y, s(z))) → cons(y, x)
cons(cons(x, z), s(y)) → transform(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.