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

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(y)

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
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))
-(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)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(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))
gcd(s(x0), 0)
gcd(0, s(x0))


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

-1(s(x), s(y)) → -1(x, y)
GCD(s(x), s(y)) → MIN(x, y)
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → -1(s(max(x, y)), s(min(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)) → 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)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(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))
gcd(s(x0), 0)
gcd(0, s(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

-1(s(x), s(y)) → -1(x, y)
GCD(s(x), s(y)) → MIN(x, y)
MIN(s(x), s(y)) → MIN(x, y)
MAX(s(x), s(y)) → MAX(x, y)
GCD(s(x), s(y)) → -1(s(max(x, y)), s(min(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)) → 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)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ UsableRulesProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

-1(s(x), s(y)) → -1(x, y)

R is empty.
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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))
gcd(s(x0), 0)
gcd(0, s(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ 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))
-(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)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP

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

MAX(s(x), s(y)) → MAX(x, y)

R is empty.
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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))
gcd(s(x0), 0)
gcd(0, s(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ 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))
-(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)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP

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

MIN(s(x), s(y)) → MIN(x, y)

R is empty.
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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))
gcd(s(x0), 0)
gcd(0, s(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ UsableRulesProof

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)))
gcd(s(x), 0) → s(x)
gcd(0, s(y)) → s(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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof

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:

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

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))
gcd(s(x0), 0)
gcd(0, s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

gcd(s(x0), s(x1))
gcd(s(x0), 0)
gcd(0, s(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ Rewriting

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:

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

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))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule GCD(s(x), s(y)) → GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y))) at position [0] we obtained the following new rules:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Rewriting
QDP
                            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

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))

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

M( max(x1, x2) ) =
/0\
\1/
+
/10\
\01/
·x1+
/10\
\01/
·x2

M( s(x1) ) =
/1\
\1/
+
/11\
\11/
·x1

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

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Rewriting
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.