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:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)



QTRS
  ↳ DependencyPairsProof

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

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)


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

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
F(true, x, y) → GT(x, y)
GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)

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:

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
F(true, x, y) → GT(x, y)
GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)

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:

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)

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:

GT(s(u), s(v)) → GT(u, v)

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

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)

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.

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)



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

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

GT(s(u), s(v)) → GT(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:

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), s(x), s(s(y)))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)

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:

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
f(true, x0, x1)
gt(s(x0), 0)

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.

f(true, 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:

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)

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

F(true, 0, x0) → F(false, s(0), s(s(x0)))
F(true, s(x0), s(x1)) → F(gt(x0, x1), s(s(x0)), s(s(s(x1))))
F(true, s(x0), 0) → F(true, s(s(x0)), s(s(0)))



↳ 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:

F(true, 0, x0) → F(false, s(0), s(s(x0)))
F(true, s(x0), 0) → F(true, s(s(x0)), s(s(0)))
F(true, s(x0), s(x1)) → F(gt(x0, x1), s(s(x0)), s(s(s(x1))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)

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

↳ 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:

F(true, s(x0), s(x1)) → F(gt(x0, x1), s(s(x0)), s(s(s(x1))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)

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

F(true, s(s(z0)), s(s(s(z1)))) → F(gt(s(z0), s(s(z1))), s(s(s(z0))), s(s(s(s(s(z1))))))



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

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

F(true, s(s(z0)), s(s(s(z1)))) → F(gt(s(z0), s(s(z1))), s(s(s(z0))), s(s(s(s(s(z1))))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)

We have to consider all minimal (P,Q,R)-chains.
By rewriting [13] the rule F(true, s(s(z0)), s(s(s(z1)))) → F(gt(s(z0), s(s(z1))), s(s(s(z0))), s(s(s(s(s(z1)))))) at position [0] we obtained the following new rules:

F(true, s(s(z0)), s(s(s(z1)))) → F(gt(z0, s(z1)), s(s(s(z0))), s(s(s(s(s(z1))))))



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

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

F(true, s(s(z0)), s(s(s(z1)))) → F(gt(z0, s(z1)), s(s(s(z0))), s(s(s(s(s(z1))))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)

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 F(true, x, y) → F(gt(x, y), s(x), s(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(F(x1, x2, x3)) = -1 - x1 + x2 - x3   
POL(c) = -1   
POL(false) = 1   
POL(gt(x1, x2)) = 0   
POL(s(x1)) = 2 + x1   
POL(true) = 0   

The following pairs are in P>:

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
The following pairs are in Pbound:

F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
The following rules are usable:

truegt(s(u), 0)
falsegt(0, v)
gt(u, v) → gt(s(u), s(v))


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

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

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

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), s(x1))
gt(s(x0), 0)

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