YES Termination w.r.t. Q proof of GTSSK07_cade10.ari

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

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

(0) Obligation:

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

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
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

trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
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), trunc(x), 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), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
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)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
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), trunc(x), s(y))
F(true, x, y) → GT(x, y)
F(true, x, y) → TRUNC(x)
TRUNC(s(s(x))) → TRUNC(x)
GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
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)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
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 3 SCCs with 2 less nodes.

(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), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
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)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
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)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
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)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
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) YES

(14) Obligation:

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

TRUNC(s(s(x))) → TRUNC(x)

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
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)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
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:

TRUNC(s(s(x))) → TRUNC(x)

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

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

(18) Obligation:

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

TRUNC(s(s(x))) → TRUNC(x)

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

(19) 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:

  • TRUNC(s(s(x))) → TRUNC(x)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

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

The TRS R consists of the following rules:

f(true, x, y) → f(gt(x, y), trunc(x), s(y))
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))
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)
trunc(0)
trunc(s(0))
trunc(s(s(x0)))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

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

(22) 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.

(23) Obligation:

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

F(true, x, y) → F(gt(x, y), trunc(x), 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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

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

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

(24) 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)

(25) Obligation:

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

F(true, x, y) → F(gt(x, y), trunc(x), 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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

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

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

(26) 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), trunc(x), s(y)) the following chains were created:
  • We consider the chain F(true, x0, x1) → F(gt(x0, x1), trunc(x0), s(x1)), F(true, x2, x3) → F(gt(x2, x3), trunc(x2), s(x3)) which results in the following constraint:
    (1)    (F(gt(x0, x1), trunc(x0), s(x1))=F(true, x2, x3) ⇒ F(true, x0, x1)≥F(gt(x0, x1), trunc(x0), s(x1)))


    We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
    (2)    (gt(x0, x1)=true ⇒ F(true, x0, x1)≥F(gt(x0, x1), trunc(x0), 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(x5), 0)≥F(gt(s(x5), 0), trunc(s(x5)), s(0)))

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


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (F(true, s(x5), 0)≥F(gt(s(x5), 0), trunc(s(x5)), s(0)))


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






To summarize, we get the following constraints P≥ for the following pairs.
  • F(true, x, y) → F(gt(x, y), trunc(x), s(y))
    • (F(true, s(x5), 0)≥F(gt(s(x5), 0), trunc(s(x5)), s(0)))
    • (F(true, x7, x6)≥F(gt(x7, x6), trunc(x7), s(x6)) ⇒ F(true, s(x7), s(x6))≥F(gt(s(x7), s(x6)), trunc(s(x7)), 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) = 1   
POL(F(x1, x2, x3)) = -1 + x1 + x2 - x3   
POL(c) = -1   
POL(false) = 0   
POL(gt(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 0   
POL(trunc(x1)) = x1   

The following pairs are in P>:

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

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

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

(27) 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)
trunc(0) → 0
trunc(s(0)) → 0
trunc(s(s(x))) → s(s(trunc(x)))

The set Q consists of the following terms:

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

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) YES