Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF(false, s(X), s(Y)) → MINUS(Y, X)
IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
MINUS(X, s(Y)) → PRED(minus(X, Y))
LE(s(X), s(Y)) → LE(X, Y)
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) → LE(Y, X)
IF(true, s(X), s(Y)) → MINUS(X, Y)
MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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

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

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

IF(false, s(X), s(Y)) → MINUS(Y, X)
IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
MINUS(X, s(Y)) → PRED(minus(X, Y))
LE(s(X), s(Y)) → LE(X, Y)
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) → LE(Y, X)
IF(true, s(X), s(Y)) → MINUS(X, Y)
MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

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

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

IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → MINUS(Y, X)
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
MINUS(X, s(Y)) → PRED(minus(X, Y))
LE(s(X), s(Y)) → LE(X, Y)
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) → LE(Y, X)
IF(true, s(X), s(Y)) → MINUS(X, Y)
MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 4 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

LE(s(X), s(Y)) → LE(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LE(s(X), s(Y)) → LE(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LE(x1, x2)  =  LE(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
s1 > LE1

Status:
LE1: [1]
s1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, 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.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MINUS(X, s(Y)) → MINUS(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
s1 > MINUS1

Status:
MINUS1: [1]
s1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, 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.

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

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

IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(x1))

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