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:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)



QTRS
  ↳ DependencyPairsProof

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

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, 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:

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → GE(x, s(y))
GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)

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:

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))
MINUS(x, y) → GE(x, s(y))
GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)

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 [13] we can delete all non-usable rules [14] from R.

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

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

GE(s(u), s(v)) → GE(u, v)

R is empty.
The set Q consists of the following terms:

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)

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.

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
QDP
                    ↳ QDPSizeChangeProof
          ↳ QDP

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

GE(s(u), s(v)) → GE(u, v)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [16] together with the size-change analysis [27] 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

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

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

minus(x, y) → cond(ge(x, s(y)), x, y)
cond(false, x, y) → 0
cond(true, x, y) → s(minus(x, s(y)))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)

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 [13] we can delete all non-usable rules [14] from R.

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

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

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
minus(x0, x1)
ge(0, s(x0))
ge(s(x0), s(x1))
cond(true, x0, x1)
cond(false, x0, x1)

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.

minus(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
QDP
                    ↳ Narrowing
                    ↳ NonInfProof

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

MINUS(x, y) → COND(ge(x, s(y)), x, y)
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [13] the rule MINUS(x, y) → COND(ge(x, s(y)), x, y) at position [0] we obtained the following new rules:

MINUS(0, x0) → COND(false, 0, x0)
MINUS(s(x0), x1) → COND(ge(x0, x1), s(x0), x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Narrowing
QDP
                        ↳ DependencyGraphProof
                    ↳ NonInfProof

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

MINUS(0, x0) → COND(false, 0, x0)
COND(true, x, y) → MINUS(x, s(y))
MINUS(s(x0), x1) → COND(ge(x0, x1), s(x0), x1)

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ DependencyGraphProof
QDP
                            ↳ Instantiation
                    ↳ NonInfProof

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

COND(true, x, y) → MINUS(x, s(y))
MINUS(s(x0), x1) → COND(ge(x0, x1), s(x0), x1)

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule MINUS(s(x0), x1) → COND(ge(x0, x1), s(x0), x1) we obtained the following new rules:

MINUS(s(x0), s(z1)) → COND(ge(x0, s(z1)), s(x0), s(z1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ Instantiation
QDP
                                ↳ Instantiation
                    ↳ NonInfProof

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

MINUS(s(x0), s(z1)) → COND(ge(x0, s(z1)), s(x0), s(z1))
COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [13] the rule COND(true, x, y) → MINUS(x, s(y)) we obtained the following new rules:

COND(true, s(z0), s(z1)) → MINUS(s(z0), s(s(z1)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ Instantiation
                              ↳ QDP
                                ↳ Instantiation
QDP
                    ↳ NonInfProof

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

MINUS(s(x0), s(z1)) → COND(ge(x0, s(z1)), s(x0), s(z1))
COND(true, s(z0), s(z1)) → MINUS(s(z0), s(s(z1)))

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus with the following steps:
Note that final constraints are written in bold face.

For Pair MINUS(x, y) → COND(ge(x, s(y)), x, y) the following chains were created:



For Pair COND(true, x, y) → MINUS(x, s(y)) the following chains were created:



To summarize, we get the following constraints P for the following pairs.
The constraints for P> respective Pbound are constructed from P where we just replace every occurence of "t ≥ s" in P by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved.
Polynomial interpretation [21]:

POL(0) = 0   
POL(COND(x1, x2, x3)) = -1 - x1 + 2·x2 - x3   
POL(MINUS(x1, x2)) = 1 + 2·x1 - x2   
POL(c) = -1   
POL(false) = 1   
POL(ge(x1, x2)) = 0   
POL(s(x1)) = 2 + x1   
POL(true) = 0   

The following pairs are in P>:

MINUS(x, y) → COND(ge(x, s(y)), x, y)
The following pairs are in Pbound:

MINUS(x, y) → COND(ge(x, s(y)), x, y)
The following rules are usable:

falsege(0, s(v))
truege(u, 0)
ge(u, v) → ge(s(u), s(v))


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QReductionProof
                  ↳ QDP
                    ↳ Narrowing
                    ↳ NonInfProof
QDP
                        ↳ DependencyGraphProof

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

COND(true, x, y) → MINUS(x, s(y))

The TRS R consists of the following rules:

ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
ge(u, 0) → true

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.