Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 AAECC Innermost (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
      ↳19 Instantiation (⇔)
        ↳20 QDP
      ↳21 NonInfProof (⇔)
        ↳22 QDP
        ↳23 PisEmptyProof (⇔)
        ↳24 TRUE

(0) Obligation:

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)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] 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)

The TRS R 2 is

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

The signature Sigma is {f}

(2) Obligation:

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:

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

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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:

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

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(11) Obligation:

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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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:

  • GT(s(u), s(v)) → GT(u, v)
    The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

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:

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

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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:

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

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(true, x0, x1)

(18) Obligation:

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

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

(19) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(true, x, y) → F(gt(x, y), s(x), s(s(y))) we obtained the following new rules [LPAR04]:

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

(20) Obligation:

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

F(true, s(z0), s(s(z1))) → F(gt(s(z0), s(s(z1))), s(s(z0)), 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), 0)
gt(s(x0), s(x1))

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

(21) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] 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:
  • We consider the chain F(true, x, y) → F(gt(x, y), s(x), s(s(y))), F(true, x, y) → F(gt(x, y), s(x), s(s(y))) which results in the following constraint:

    (1)    (F(gt(x0, x1), s(x0), s(s(x1)))=F(true, x2, x3) ⇒ F(true, x2, x3)≥F(gt(x2, x3), s(x2), s(s(x3))))



    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:

    (2)    (gt(x0, x1)=true ⇒ F(true, s(x0), s(s(x1)))≥F(gt(s(x0), s(s(x1))), s(s(x0)), s(s(s(s(x1))))))



    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x0, x1)=true which results in the following new constraints:

    (3)    (true=true ⇒ F(true, s(s(x5)), s(s(0)))≥F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0))))))


    (4)    (gt(x7, x6)=true∧(gt(x7, x6)=true ⇒ F(true, s(x7), s(s(x6)))≥F(gt(s(x7), s(s(x6))), s(s(x7)), s(s(s(s(x6)))))) ⇒ F(true, s(s(x7)), s(s(s(x6))))≥F(gt(s(s(x7)), s(s(s(x6)))), s(s(s(x7))), s(s(s(s(s(x6)))))))



    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:

    (5)    (F(true, s(s(x5)), s(s(0)))≥F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0))))))



    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gt(x7, x6)=true ⇒ F(true, s(x7), s(s(x6)))≥F(gt(s(x7), s(s(x6))), s(s(x7)), s(s(s(s(x6)))))) with σ = [ ] which results in the following new constraint:

    (6)    (F(true, s(x7), s(s(x6)))≥F(gt(s(x7), s(s(x6))), s(s(x7)), s(s(s(s(x6))))) ⇒ F(true, s(s(x7)), s(s(s(x6))))≥F(gt(s(s(x7)), s(s(s(x6)))), s(s(s(x7))), s(s(s(s(s(x6)))))))







To summarize, we get the following constraints P≥ for the following pairs.
  • F(true, x, y) → F(gt(x, y), s(x), s(s(y)))
    • (F(true, s(s(x5)), s(s(0)))≥F(gt(s(s(x5)), s(s(0))), s(s(s(x5))), s(s(s(s(0))))))
    • (F(true, s(x7), s(s(x6)))≥F(gt(s(x7), s(s(x6))), s(s(x7)), s(s(s(s(x6))))) ⇒ F(true, s(s(x7)), s(s(s(x6))))≥F(gt(s(s(x7)), s(s(s(x6)))), s(s(s(x7))), s(s(s(s(s(x6)))))))




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 [NONINF]:

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

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:

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

(22) Obligation:

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

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(24) TRUE