YES Termination w.r.t. Q proof of GTSSK07_cade13t.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 6 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 NonInfProof (⇔, 0 ms)
        ↳27 AND
            ↳28 QDP
                ↳29 DependencyGraphProof (⇔, 0 ms)
                ↳30 TRUE
            ↳31 QDP
                ↳32 DependencyGraphProof (⇔, 0 ms)
                ↳33 TRUE

(0) Obligation:

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

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(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:

DIV(x, s(y)) → D(x, s(y), 0)
D(x, s(y), z) → COND(ge(x, z), x, y, z)
D(x, s(y), z) → GE(x, z)
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
COND(true, x, y, z) → PLUS(s(y), z)
GE(s(u), s(v)) → GE(u, v)
PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(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 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(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:

PLUS(n, s(m)) → PLUS(n, m)

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

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(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].

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

(11) Obligation:

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

PLUS(n, s(m)) → PLUS(n, m)

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(n, s(m)) → PLUS(n, m)
    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:

GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(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:

GE(s(u), s(v)) → GE(u, v)

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

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(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].

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

(18) 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.

(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:

  • GE(s(u), s(v)) → GE(u, v)
    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, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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].

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)

(25) Obligation:

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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, z) → D(x, s(y), plus(s(y), z)) the following chains were created:
  • We consider the chain D(x3, s(x4), x5) → COND(ge(x3, x5), x3, x4, x5), COND(true, x6, x7, x8) → D(x6, s(x7), plus(s(x7), x8)) which results in the following constraint:
    (1)    (COND(ge(x3, x5), x3, x4, x5)=COND(true, x6, x7, x8) ⇒ COND(true, x6, x7, x8)≥D(x6, s(x7), plus(s(x7), x8)))


    We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
    (2)    (ge(x3, x5)=true ⇒ COND(true, x3, x4, x5)≥D(x3, s(x4), plus(s(x4), x5)))


    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x3, x5)=true which results in the following new constraints:
    (3)    (true=true ⇒ COND(true, x18, x4, 0)≥D(x18, s(x4), plus(s(x4), 0)))

    (4)    (ge(x21, x20)=true∧(∀x22:ge(x21, x20)=true ⇒ COND(true, x21, x22, x20)≥D(x21, s(x22), plus(s(x22), x20))) ⇒ COND(true, s(x21), x4, s(x20))≥D(s(x21), s(x4), plus(s(x4), s(x20))))


    We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
    (5)    (COND(true, x18, x4, 0)≥D(x18, s(x4), plus(s(x4), 0)))


    We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x22:ge(x21, x20)=true ⇒ COND(true, x21, x22, x20)≥D(x21, s(x22), plus(s(x22), x20))) with σ = [x22 / x4] which results in the following new constraint:
    (6)    (COND(true, x21, x4, x20)≥D(x21, s(x4), plus(s(x4), x20)) ⇒ COND(true, s(x21), x4, s(x20))≥D(s(x21), s(x4), plus(s(x4), s(x20))))






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


    We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:
    (2)    (D(x9, s(x10), x14)≥COND(ge(x9, x14), x9, x10, x14))






To summarize, we get the following constraints P≥ for the following pairs.
  • COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
    • (COND(true, x18, x4, 0)≥D(x18, s(x4), plus(s(x4), 0)))
    • (COND(true, x21, x4, x20)≥D(x21, s(x4), plus(s(x4), x20)) ⇒ COND(true, s(x21), x4, s(x20))≥D(s(x21), s(x4), plus(s(x4), s(x20))))

  • D(x, s(y), z) → COND(ge(x, z), x, y, z)
    • (D(x9, s(x10), x14)≥COND(ge(x9, x14), x9, x10, 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) = 0   
POL(COND(x1, x2, x3, x4)) = x2 - x4   
POL(D(x1, x2, x3)) = x1 + x2 - x3   
POL(c) = -1   
POL(false) = 1   
POL(ge(x1, x2)) = 1 + x1   
POL(plus(x1, x2)) = x1 + x2   
POL(s(x1)) = 1 + x1   
POL(true) = 1   

The following pairs are in P>:

D(x, s(y), z) → COND(ge(x, z), x, y, z)
The following pairs are in Pbound:

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
The following rules are usable:

n → plus(n, 0)
s(plus(n, m)) → plus(n, s(m))

(27) Complex Obligation (AND)

(28) Obligation:

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))

The TRS R consists of the following rules:

ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

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

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

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) TRUE

(31) Obligation:

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

D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

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

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

(32) DependencyGraphProof (EQUIVALENT transformation)

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

(33) TRUE