Termination w.r.t. Q of the given QTRS could be proven:

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

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)
-¹(s(x), s(y)) → -¹(x, y)
GCD(s(x), s(y), z) → GCD(-(max(x, y), min(x, y)), s(min(x, y)), z)
GCD(s(x), s(y), z) → -¹(max(x, y), min(x, y))
GCD(s(x), s(y), z) → MAX(x, y)
GCD(s(x), s(y), z) → MIN(x, y)
GCD(x, s(y), s(z)) → GCD(x, -(max(y, z), min(y, z)), s(min(y, z)))
GCD(x, s(y), s(z)) → -¹(max(y, z), min(y, z))
GCD(x, s(y), s(z)) → MAX(y, z)
GCD(x, s(y), s(z)) → MIN(y, z)
GCD(s(x), y, s(z)) → GCD(-(max(x, z), min(x, z)), y, s(min(x, z)))
GCD(s(x), y, s(z)) → -¹(max(x, z), min(x, z))
GCD(s(x), y, s(z)) → MAX(x, z)
GCD(s(x), y, s(z)) → MIN(x, z)

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

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 4 SCCs with 9 less nodes.

(4) Complex Obligation (AND)

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

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

-¹(s(x), s(y)) → -¹(x, y)

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(9) TRUE

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

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

MAX(s(x), s(y)) → MAX(x, y)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(14) TRUE

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

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

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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

MIN(s(x), s(y)) → MIN(x, y)

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

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

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

(19) TRUE

(20) Obligation:

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

GCD(x, s(y), s(z)) → GCD(x, -(max(y, z), min(y, z)), s(min(y, z)))
GCD(s(x), s(y), z) → GCD(-(max(x, y), min(x, y)), s(min(x, y)), z)
GCD(s(x), y, s(z)) → GCD(-(max(x, z), min(x, z)), y, s(min(x, z)))

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

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.

GCD(x, s(y), s(z)) → GCD(x, -(max(y, z), min(y, z)), s(min(y, z)))
GCD(s(x), s(y), z) → GCD(-(max(x, y), min(x, y)), s(min(x, y)), z)
GCD(s(x), y, s(z)) → GCD(-(max(x, z), min(x, z)), y, s(min(x, z)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(GCD(x₁, x₂, x₃)) = (3/4)x₁ + (1/2)x₂ + (1/4)x₃   
POL(s(x₁)) = 1 + (3)x₁   
POL(-(x₁, x₂)) = 1/4 + x₁   
POL(max(x₁, x₂)) = x₁ + x₂   
POL(min(x₁, x₂)) = x₁   
POL(0) = 0   

The value of delta used in the strict ordering is 3/8.
The following usable rules [FROCOS05] were oriented:

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

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

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