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

Q is empty.


QTRS
  ↳ AAECC Innermost

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

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

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

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

The TRS R 2 is

f(true, x, y, z) → f(and(gt(x, y), gt(x, z)), x, s(y), z)
f(true, x, y, z) → f(and(gt(x, y), gt(x, z)), x, y, s(z))

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)


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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ AAECC Innermost
    ↳ 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, z) → f(and(gt(x, y), gt(x, z)), x, s(y), z)
f(true, x, y, z) → f(and(gt(x, y), gt(x, z)), x, y, s(z))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)
and(x, true) → x
and(x, false) → false

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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
  ↳ AAECC Innermost
    ↳ 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:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)



↳ QTRS
  ↳ AAECC Innermost
    ↳ 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 [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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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
  ↳ AAECC Innermost
    ↳ 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, z) → F(and(gt(x, y), gt(x, z)), x, s(y), z)
F(true, x, y, z) → F(and(gt(x, y), gt(x, z)), x, y, s(z))

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)
and(x, true) → x
and(x, false) → false

The set Q consists of the following terms:

f(true, x0, x1, x2)
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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, x2)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ NonInfProof

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

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

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)
and(x, true) → x
and(x, false) → false

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus [18] with the following steps:
Note that final constraints are written in bold face.


For Pair F(true, x, y, z) → F(and(gt(x, y), gt(x, z)), x, s(y), z) the following chains were created:




For Pair F(true, x, y, z) → F(and(gt(x, y), gt(x, z)), x, y, s(z)) 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 [18]:

POL(0) = 0   
POL(F(x1, x2, x3, x4)) = -1 - x1 + x2 - x4   
POL(and(x1, x2)) = 0   
POL(c) = -1   
POL(false) = 1   
POL(gt(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

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

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

falseand(x, false)
xand(x, true)


↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ NonInfProof
QDP
                            ↳ NonInfProof

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

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

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)
and(x, true) → x
and(x, false) → false

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

We have to consider all minimal (P,Q,R)-chains.
The DP Problem is simplified using the Induction Calculus [18] with the following steps:
Note that final constraints are written in bold face.


For Pair F(true, x, y, z) → F(and(gt(x, y), gt(x, z)), x, s(y), z) 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 [18]:

POL(0) = 0   
POL(F(x1, x2, x3, x4)) = -1 - x1 + x2 - x3 + x4   
POL(and(x1, x2)) = 0   
POL(c) = -1   
POL(false) = 1   
POL(gt(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

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

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

falseand(x, false)
xand(x, true)


↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ NonInfProof
                          ↳ QDP
                            ↳ 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)
and(x, true) → x
and(x, false) → false

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
and(x0, true)
and(x0, false)

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