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

(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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
min(x, 0) → 0

The TRS R 2 is

gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The signature Sigma is {gcd}

(2) 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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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)
-¹(s(x), s(y)) → -¹(x, y)
GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
GCD(s(x), s(y)) → -¹(s(max(x, y)), s(min(x, y)))
GCD(s(x), s(y)) → MAX(x, y)
GCD(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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

-¹(s(x), s(y)) → -¹(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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

-¹(s(x), s(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2)  =  -¹(x1)
s(x1)  =  s(x1)
min(x1, x2)  =  x2
0  =  0
max(x1, x2)  =  max(x1, x2)
-(x1, x2)  =  -(x1)
gcd(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:

max₂ > s₁ > -^1₁ > 0
-₁ > 0

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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

(9) 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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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

(13) 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(x1)
s(x1)  =  s(x1)
min(x1, x2)  =  x2
0  =  0
max(x1, x2)  =  max(x1, x2)
-(x1, x2)  =  -(x1)
gcd(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:

max₂ > s₁ > MAX₁ > 0
-₁ > 0

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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

(14) 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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) 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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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

(18) 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(x1)
s(x1)  =  s(x1)
min(x1, x2)  =  x2
0  =  0
max(x1, x2)  =  max(x1, x2)
-(x1, x2)  =  -(x1)
gcd(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:

max₂ > s₁ > MIN₁ > 0
-₁ > 0

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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

(19) 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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(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))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
gcd(s(x), s(y)) → gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

The set Q consists of the following terms:

min(x0, 0)
min(0, x0)
min(s(x0), s(x1))
max(x0, 0)
max(0, x0)
max(s(x0), s(x1))
-(x0, 0)
-(s(x0), s(x1))
gcd(s(x0), s(x1))

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