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, 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.


↳ 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, 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.

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)) → MIN(x, y)
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
MINUS(s(x), s(y)) → ANY(y)
MIN(s(x), s(y)) → MIN(x, y)
GCD(s(x), s(y)) → MINUS(max(x, y), min(x, y))
ANY(s(x)) → ANY(x)
MAX(s(x), s(y)) → MAX(x, y)
MINUS(s(x), s(y)) → MINUS(x, any(y))
GCD(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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

GCD(s(x), s(y)) → MIN(x, y)
GCD(s(x), s(y)) → GCD(minus(max(x, y), min(x, y)), s(min(x, y)))
MINUS(s(x), s(y)) → ANY(y)
MIN(s(x), s(y)) → MIN(x, y)
GCD(s(x), s(y)) → MINUS(max(x, y), min(x, y))
ANY(s(x)) → ANY(x)
MAX(s(x), s(y)) → MAX(x, y)
MINUS(s(x), s(y)) → MINUS(x, any(y))
GCD(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.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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.
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
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ANY(s(x)) → ANY(x)

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: 



↳ 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, 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.
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, any(y))

The TRS R consists of the following rules:

any(s(x)) → s(s(any(x)))
any(x) → x

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: 



↳ 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, 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.
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
          ↳ 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)

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: 



↳ 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, 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.
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
          ↳ 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)

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: 



↳ 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, 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.
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, y)), s(min(x, y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( minus(x1, x2) ) =
/0\
\0/
+
/10\
\01/
·x1+
/00\
\00/
·x2

M( max(x1, x2) ) =
/0\
\0/
+
/10\
\01/
·x1+
/10\
\01/
·x2

M( any(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( s(x1) ) =
/0\
\1/
+
/11\
\11/
·x1

M( min(x1, x2) ) =
/0\
\0/
+
/10\
\01/
·x1+
/00\
\00/
·x2

M( 0 ) =
/0\
\0/

Tuple symbols:
M( GCD(x1, x2) ) = 0+
[1,1]
·x1+
[1,0]
·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:

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



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