YES Termination w.r.t. Q proof of AProVE_07_thiemann27.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 (⇔, 5 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

LE(s(x), s(y)) → LE(x, y)
INT(x, y) → IF(le(x, y), x, y)
INT(x, y) → LE(x, y)
IF(true, x, y) → INT(s(x), y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(11) Obligation:

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

LE(s(x), s(y)) → LE(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:

  • LE(s(x), s(y)) → LE(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:

INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
int(x, y) → if(le(x, y), x, y)
if(true, x, y) → cons(x, int(s(x), y))
if(false, x, y) → nil

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, 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:

INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
int(x0, x1)
if(true, x0, x1)
if(false, x0, 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].

int(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(18) Obligation:

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

INT(x, y) → IF(le(x, y), x, y)
IF(true, x, y) → INT(s(x), y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(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 INT(x, y) → IF(le(x, y), x, y) the following chains were created:
  • We consider the chain IF(true, x2, x3) → INT(s(x2), x3), INT(x4, x5) → IF(le(x4, x5), x4, x5) which results in the following constraint:
    (1)    (INT(s(x2), x3)=INT(x4, x5) ⇒ INT(x4, x5)≥IF(le(x4, x5), x4, x5))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (INT(s(x2), x3)≥IF(le(s(x2), x3), s(x2), x3))






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


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (le(x6, x7)=true ⇒ IF(true, x6, x7)≥INT(s(x6), x7))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x6, x7)=true which results in the following new constraints:
    (3)    (true=true ⇒ IF(true, 0, x12)≥INT(s(0), x12))

    (4)    (le(x15, x14)=true∧(le(x15, x14)=true ⇒ IF(true, x15, x14)≥INT(s(x15), x14)) ⇒ IF(true, s(x15), s(x14))≥INT(s(s(x15)), s(x14)))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (IF(true, 0, x12)≥INT(s(0), x12))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (le(x15, x14)=true ⇒ IF(true, x15, x14)≥INT(s(x15), x14)) with σ = [ ] which results in the following new constraint:
    (6)    (IF(true, x15, x14)≥INT(s(x15), x14) ⇒ IF(true, s(x15), s(x14))≥INT(s(s(x15)), s(x14)))






To summarize, we get the following constraints P≥ for the following pairs.
  • INT(x, y) → IF(le(x, y), x, y)
    • (INT(s(x2), x3)≥IF(le(s(x2), x3), s(x2), x3))

  • IF(true, x, y) → INT(s(x), y)
    • (IF(true, 0, x12)≥INT(s(0), x12))
    • (IF(true, x15, x14)≥INT(s(x15), x14) ⇒ IF(true, s(x15), s(x14))≥INT(s(s(x15)), s(x14)))




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(IF(x1, x2, x3)) = -x2 + x3   
POL(INT(x1, x2)) = -x1 + x2   
POL(c) = -1   
POL(false) = 1   
POL(le(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

IF(true, x, y) → INT(s(x), y)
The following pairs are in Pbound:

IF(true, x, y) → INT(s(x), y)
There are no usable rules

(20) Obligation:

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

INT(x, y) → IF(le(x, y), x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(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