YES Termination w.r.t. Q proof of GTSSK07_cade03.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 12 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 NonInfProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 TRUE

(0) Obligation:

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)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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:

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

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

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

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:

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

GE(s(u), s(v)) → GE(u, v)

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

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

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

(11) Obligation:

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.

(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:

  • GE(s(u), s(v)) → GE(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:

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

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

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

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

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

(18) Obligation:

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

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

(19) 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 COND(true, x, y) → MINUS(x, s(y)) the following chains were created:
  • We consider the chain MINUS(x2, x3) → COND(ge(x2, s(x3)), x2, x3), COND(true, x4, x5) → MINUS(x4, s(x5)) which results in the following constraint:
    (1)    (COND(ge(x2, s(x3)), x2, x3)=COND(true, x4, x5) ⇒ COND(true, x4, x5)≥MINUS(x4, s(x5)))


    We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:
    (2)    (s(x3)=x12∧ge(x2, x12)=true ⇒ COND(true, x2, x3)≥MINUS(x2, s(x3)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x2, x12)=true which results in the following new constraints:
    (3)    (ge(x15, x14)=true∧s(x3)=s(x14)∧(∀x16:ge(x15, x14)=true∧s(x16)=x14 ⇒ COND(true, x15, x16)≥MINUS(x15, s(x16))) ⇒ COND(true, s(x15), x3)≥MINUS(s(x15), s(x3)))

    (4)    (true=true∧s(x3)=0 ⇒ COND(true, x17, x3)≥MINUS(x17, s(x3)))


    We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:
    (5)    (ge(x15, x14)=true ⇒ COND(true, s(x15), x14)≥MINUS(s(x15), s(x14)))


    We solved constraint (4) using rules (I), (II).We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on ge(x15, x14)=true which results in the following new constraints:
    (6)    (ge(x20, x19)=true∧(ge(x20, x19)=true ⇒ COND(true, s(x20), x19)≥MINUS(s(x20), s(x19))) ⇒ COND(true, s(s(x20)), s(x19))≥MINUS(s(s(x20)), s(s(x19))))

    (7)    (true=true ⇒ COND(true, s(x21), 0)≥MINUS(s(x21), s(0)))


    We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (ge(x20, x19)=true ⇒ COND(true, s(x20), x19)≥MINUS(s(x20), s(x19))) with σ = [ ] which results in the following new constraint:
    (8)    (COND(true, s(x20), x19)≥MINUS(s(x20), s(x19)) ⇒ COND(true, s(s(x20)), s(x19))≥MINUS(s(s(x20)), s(s(x19))))


    We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
    (9)    (COND(true, s(x21), 0)≥MINUS(s(x21), s(0)))






For Pair MINUS(x, y) → COND(ge(x, s(y)), x, y) the following chains were created:
  • We consider the chain COND(true, x6, x7) → MINUS(x6, s(x7)), MINUS(x8, x9) → COND(ge(x8, s(x9)), x8, x9) which results in the following constraint:
    (1)    (MINUS(x6, s(x7))=MINUS(x8, x9) ⇒ MINUS(x8, x9)≥COND(ge(x8, s(x9)), x8, x9))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (MINUS(x6, s(x7))≥COND(ge(x6, s(s(x7))), x6, s(x7)))






To summarize, we get the following constraints P≥ for the following pairs.
  • COND(true, x, y) → MINUS(x, s(y))
    • (COND(true, s(x20), x19)≥MINUS(s(x20), s(x19)) ⇒ COND(true, s(s(x20)), s(x19))≥MINUS(s(s(x20)), s(s(x19))))
    • (COND(true, s(x21), 0)≥MINUS(s(x21), s(0)))

  • MINUS(x, y) → COND(ge(x, s(y)), x, y)
    • (MINUS(x6, s(x7))≥COND(ge(x6, s(s(x7))), x6, s(x7)))




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

The following pairs are in P>:

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

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

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

(20) Obligation:

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

MINUS(x, y) → COND(ge(x, s(y)), x, 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.

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(22) TRUE