YES Termination w.r.t. Q proof of AProVE_07_otto09.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 (⇔, 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 TransformationProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 NonInfProof (⇔, 0 ms)
        ↳29 AND
            ↳30 QDP
                ↳31 DependencyGraphProof (⇔, 0 ms)
                ↳32 TRUE
            ↳33 QDP
                ↳34 DependencyGraphProof (⇔, 0 ms)
                ↳35 TRUE

(0) Obligation:

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

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

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:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(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:

LT(s(x), s(y)) → LT(x, y)
PLUS(x, s(y)) → PLUS(x, y)
QUOT(x, s(y)) → HELP(x, s(y), 0)
HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
HELP(x, s(y), c) → LT(c, x)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
IF(true, x, s(y), c) → PLUS(c, s(y))

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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:

PLUS(x, s(y)) → PLUS(x, y)

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

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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].

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(11) Obligation:

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

PLUS(x, s(y)) → PLUS(x, y)

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:

  • PLUS(x, s(y)) → PLUS(x, y)
    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:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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:

LT(s(x), s(y)) → LT(x, y)

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

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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].

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(18) Obligation:

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

LT(s(x), s(y)) → LT(x, y)

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:

  • LT(s(x), s(y)) → LT(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(20) YES

(21) Obligation:

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

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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:

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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].

quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(25) Obligation:

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

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

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

(26) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y))) at position [2] we obtained the following new rules [LPAR04]:

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y))) → IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

(27) Obligation:

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

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

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

(28) 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 HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c) the following chains were created:
  • We consider the chain IF(true, x3, s(x4), x5) → HELP(x3, s(x4), s(plus(x5, x4))), HELP(x6, s(x7), x8) → IF(lt(x8, x6), x6, s(x7), x8) which results in the following constraint:
    (1)    (HELP(x3, s(x4), s(plus(x5, x4)))=HELP(x6, s(x7), x8) ⇒ HELP(x6, s(x7), x8)≥IF(lt(x8, x6), x6, s(x7), x8))


    We simplified constraint (1) using rules (I), (II), (III), (IV) which results in the following new constraint:
    (2)    (HELP(x3, s(x4), x8)≥IF(lt(x8, x3), x3, s(x4), x8))






For Pair IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y))) the following chains were created:
  • We consider the chain HELP(x9, s(x10), x11) → IF(lt(x11, x9), x9, s(x10), x11), IF(true, x12, s(x13), x14) → HELP(x12, s(x13), s(plus(x14, x13))) which results in the following constraint:
    (1)    (IF(lt(x11, x9), x9, s(x10), x11)=IF(true, x12, s(x13), x14) ⇒ IF(true, x12, s(x13), x14)≥HELP(x12, s(x13), s(plus(x14, x13))))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (lt(x11, x9)=true ⇒ IF(true, x9, s(x10), x11)≥HELP(x9, s(x10), s(plus(x11, x10))))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on lt(x11, x9)=true which results in the following new constraints:
    (3)    (true=true ⇒ IF(true, s(x19), s(x10), 0)≥HELP(s(x19), s(x10), s(plus(0, x10))))

    (4)    (lt(x21, x20)=true∧(∀x22:lt(x21, x20)=true ⇒ IF(true, x20, s(x22), x21)≥HELP(x20, s(x22), s(plus(x21, x22)))) ⇒ IF(true, s(x20), s(x10), s(x21))≥HELP(s(x20), s(x10), s(plus(s(x21), x10))))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (IF(true, s(x19), s(x10), 0)≥HELP(s(x19), s(x10), s(plus(0, x10))))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x22:lt(x21, x20)=true ⇒ IF(true, x20, s(x22), x21)≥HELP(x20, s(x22), s(plus(x21, x22)))) with σ = [x22 / x10] which results in the following new constraint:
    (6)    (IF(true, x20, s(x10), x21)≥HELP(x20, s(x10), s(plus(x21, x10))) ⇒ IF(true, s(x20), s(x10), s(x21))≥HELP(s(x20), s(x10), s(plus(s(x21), x10))))






To summarize, we get the following constraints P≥ for the following pairs.
  • HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
    • (HELP(x3, s(x4), x8)≥IF(lt(x8, x3), x3, s(x4), x8))

  • IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))
    • (IF(true, s(x19), s(x10), 0)≥HELP(s(x19), s(x10), s(plus(0, x10))))
    • (IF(true, x20, s(x10), x21)≥HELP(x20, s(x10), s(plus(x21, x10))) ⇒ IF(true, s(x20), s(x10), s(x21))≥HELP(s(x20), s(x10), s(plus(s(x21), x10))))




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(HELP(x1, x2, x3)) = x1 + x2 - x3   
POL(IF(x1, x2, x3, x4)) = -1 + x1 + x2 - x4   
POL(c) = -1   
POL(false) = 0   
POL(lt(x1, x2)) = 1   
POL(plus(x1, x2)) = x1 + x2   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

The following pairs are in P>:

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
The following pairs are in Pbound:

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))
The following rules are usable:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
s(plus(x, y)) → plus(x, s(y))
x → plus(x, 0)

(29) Complex Obligation (AND)

(30) Obligation:

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

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

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

(31) DependencyGraphProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

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

(34) DependencyGraphProof (EQUIVALENT transformation)

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

(35) TRUE