Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
  ↳6 QDP
    ↳7 QDPOrderProof (⇔)
    ↳8 QDP
    ↳9 PisEmptyProof (⇔)
    ↳10 TRUE
  ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 PisEmptyProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE
  ↳21 QDP

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

(6) Obligation:

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.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(s(x), s(y)) → MINUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
s1 > MINUS1

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE

(11) Obligation:

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)

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MAX(s(x), s(y)) → MAX(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MAX(x1, x2)  =  MAX(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
s1 > MAX1

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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.

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) TRUE

(16) Obligation:

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)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MIN(s(x), s(y)) → MIN(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2)  =  MIN(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
s1 > MIN1

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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.

(19) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(20) TRUE

(21) Obligation:

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.