YES Termination w.r.t. Q proof of Beerendonk_07_3.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 3 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 (⇔, 14 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES

(0) Obligation:

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

cond(true, x, y) → cond(gr(x, y), x, add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The TRS R 2 is

cond(true, x, y) → cond(gr(x, y), x, add(x, y))

The signature Sigma is {cond}

(2) Obligation:

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

cond(true, x, y) → cond(gr(x, y), x, add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(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:

COND(true, x, y) → COND(gr(x, y), x, add(x, y))
COND(true, x, y) → GR(x, y)
COND(true, x, y) → ADD(x, y)
GR(s(x), s(y)) → GR(x, y)
ADD(s(x), y) → ADD(x, y)

The TRS R consists of the following rules:

cond(true, x, y) → cond(gr(x, y), x, add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(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 3 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

ADD(s(x), y) → ADD(x, y)

The TRS R consists of the following rules:

cond(true, x, y) → cond(gr(x, y), x, add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(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:

ADD(s(x), y) → ADD(x, y)

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

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(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].

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)

(11) Obligation:

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

ADD(s(x), y) → ADD(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:

  • ADD(s(x), y) → ADD(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:

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x, y) → cond(gr(x, y), x, add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(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:

GR(s(x), s(y)) → GR(x, y)

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

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(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].

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), x1)

(18) Obligation:

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

GR(s(x), s(y)) → GR(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:

  • GR(s(x), s(y)) → GR(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:

COND(true, x, y) → COND(gr(x, y), x, add(x, y))

The TRS R consists of the following rules:

cond(true, x, y) → cond(gr(x, y), x, add(x, y))
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), 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:

COND(true, x, y) → COND(gr(x, y), x, add(x, y))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), 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].

cond(true, x0, x1)

(25) Obligation:

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

COND(true, x, y) → COND(gr(x, y), x, add(x, y))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), 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 COND(true, x, y) → COND(gr(x, y), x, add(x, y)) the following chains were created:
  • We consider the chain COND(true, x0, x1) → COND(gr(x0, x1), x0, add(x0, x1)), COND(true, x2, x3) → COND(gr(x2, x3), x2, add(x2, x3)) which results in the following constraint:
    (1)    (COND(gr(x0, x1), x0, add(x0, x1))=COND(true, x2, x3) ⇒ COND(true, x0, x1)≥COND(gr(x0, x1), x0, add(x0, x1)))


    We simplified constraint (1) using rules (I), (II), (IV) which results in the following new constraint:
    (2)    (gr(x0, x1)=true ⇒ COND(true, x0, x1)≥COND(gr(x0, x1), x0, add(x0, x1)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x0, x1)=true which results in the following new constraints:
    (3)    (true=true ⇒ COND(true, s(x5), 0)≥COND(gr(s(x5), 0), s(x5), add(s(x5), 0)))

    (4)    (gr(x7, x6)=true∧(gr(x7, x6)=true ⇒ COND(true, x7, x6)≥COND(gr(x7, x6), x7, add(x7, x6))) ⇒ COND(true, s(x7), s(x6))≥COND(gr(s(x7), s(x6)), s(x7), add(s(x7), s(x6))))


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


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






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

The following pairs are in P>:

COND(true, x, y) → COND(gr(x, y), x, add(x, y))
The following pairs are in Pbound:

COND(true, x, y) → COND(gr(x, y), x, add(x, y))
The following rules are usable:

false → gr(0, x)
true → gr(s(x), 0)
gr(x, y) → gr(s(x), s(y))
x → add(0, x)
s(add(x, y)) → add(s(x), y)

(27) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
add(0, x) → x
add(s(x), y) → s(add(x, y))

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
add(0, x0)
add(s(x0), 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