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 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE
  ↳25 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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → 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, any(y))
MINUS(s(x), s(y)) → ANY(y)
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
GCD(s(x), s(y)) → MINUS(max(x, y), min(x, y))
GCD(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → MIN(x, y)
ANY(s(x)) → ANY(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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → 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 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ANY(s(x)) → ANY(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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ANY(s(x)) → ANY(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ANY(x1)  =  ANY(x1)
s(x1)  =  s(x1)
min(x1, x2)  =  min(x1, x2)
0  =  0
max(x1, x2)  =  max(x1, x2)
minus(x1, x2)  =  x1
any(x1)  =  any(x1)
gcd(x1, x2)  =  gcd

Lexicographic Path Order [LPO].
Precedence:
min2 > [s1, gcd]
min2 > 0
max2 > [s1, gcd]
any1 > [s1, gcd]


The following usable rules [FROCOS05] were oriented:

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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

(7) 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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, any(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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(11) 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, any(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x1)
s(x1)  =  s(x1)
any(x1)  =  any(x1)
min(x1, x2)  =  min(x1, x2)
0  =  0
max(x1, x2)  =  max(x1, x2)
minus(x1, x2)  =  x1
gcd(x1, x2)  =  gcd

Lexicographic Path Order [LPO].
Precedence:
[MINUS1, any1] > [s1, gcd]
min2 > [s1, gcd]
min2 > 0
max2 > [s1, gcd]


The following usable rules [FROCOS05] were oriented:

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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

(12) 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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(16) 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)  =  x2
s(x1)  =  s(x1)
min(x1, x2)  =  min(x1, x2)
0  =  0
max(x1, x2)  =  max(x1, x2)
minus(x1, x2)  =  x1
any(x1)  =  any(x1)
gcd(x1, x2)  =  gcd

Lexicographic Path Order [LPO].
Precedence:
min2 > [s1, gcd]
min2 > 0
max2 > [s1, gcd]
any1 > [s1, gcd]


The following usable rules [FROCOS05] were oriented:

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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

(17) 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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(21) 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)  =  x2
s(x1)  =  s(x1)
min(x1, x2)  =  min(x1, x2)
0  =  0
max(x1, x2)  =  max(x1, x2)
minus(x1, x2)  =  x1
any(x1)  =  any(x1)
gcd(x1, x2)  =  gcd

Lexicographic Path Order [LPO].
Precedence:
min2 > [s1, gcd]
min2 > 0
max2 > [s1, gcd]
any1 > [s1, gcd]


The following usable rules [FROCOS05] were oriented:

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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

(22) 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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE

(25) Obligation:

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

GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, 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, any(y)))
gcd(s(x), s(y)) → gcd(minus(max(x, y), min(x, y)), s(min(x, y)))
any(s(x)) → s(s(any(x)))
any(x) → x

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