YES Termination w.r.t. Q proof of AProVE_07_thiemann23.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 RemovalProof (⇒, 0 ms)
        ↳20 QDP
        ↳21 NonInfProof (⇔, 8 ms)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔, 0 ms)
        ↳24 TRUE

(0) Obligation:

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

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

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:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
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:

digits
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(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:

DIGITS → D(0)
D(x) → IF(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
D(x) → LE(x, s(s(s(s(s(s(s(s(s(0))))))))))
IF(true, x) → D(s(x))
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
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:

digits
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(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 2 less nodes.

(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:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
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:

digits
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(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:

LE(s(x), s(y)) → LE(x, y)

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

digits
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(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].

digits
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(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:

IF(true, x) → D(s(x))
D(x) → IF(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)

The TRS R consists of the following rules:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
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:

digits
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(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:

IF(true, x) → D(s(x))
D(x) → IF(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

digits
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(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].

digits
d(x0)
if(true, x0)
if(false, x0)

(18) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)

The TRS R consists of the following rules:

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

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) RemovalProof (SOUND transformation)

In the following pairs the term without variables s(s(s(s(s(s(s(s(s(0))))))))) is replaced by the fresh variable x_removed.
Pair: D(x) → IF(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
Positions in right side of the pair:
  • [0,1]
The new variable was added to all pairs as a new argument[CONREM].

(20) Obligation:

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

IF(true, x, x_removed) → D(s(x), x_removed)
D(x, x_removed) → IF(le(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

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

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


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


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x2, x3)=true which results in the following new constraints:
    (3)    (true=true ⇒ IF(true, 0, x12)≥D(s(0), x12))

    (4)    (le(x14, x13)=true∧(le(x14, x13)=true ⇒ IF(true, x14, x13)≥D(s(x14), x13)) ⇒ IF(true, s(x14), s(x13))≥D(s(s(x14)), s(x13)))


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


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (le(x14, x13)=true ⇒ IF(true, x14, x13)≥D(s(x14), x13)) with σ = [ ] which results in the following new constraint:
    (6)    (IF(true, x14, x13)≥D(s(x14), x13) ⇒ IF(true, s(x14), s(x13))≥D(s(s(x14)), s(x13)))






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


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






To summarize, we get the following constraints P≥ for the following pairs.
  • IF(true, x, x_removed) → D(s(x), x_removed)
    • (IF(true, 0, x12)≥D(s(0), x12))
    • (IF(true, x14, x13)≥D(s(x14), x13) ⇒ IF(true, s(x14), s(x13))≥D(s(s(x14)), s(x13)))

  • D(x, x_removed) → IF(le(x, x_removed), x, x_removed)
    • (D(s(x6), x7)≥IF(le(s(x6), x7), s(x6), 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(D(x1, x2)) = -1 - x1 + x2   
POL(IF(x1, x2, x3)) = -1 - x2 + x3   
POL(c) = -2   
POL(false) = 1   
POL(le(x1, x2)) = 1 + x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

The following pairs are in P>:

IF(true, x, x_removed) → D(s(x), x_removed)
The following pairs are in Pbound:

IF(true, x, x_removed) → D(s(x), x_removed)
There are no usable rules

(22) Obligation:

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

D(x, x_removed) → IF(le(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

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

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.

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) TRUE