YES Termination w.r.t. Q proof of AProVE_07_thiemann10.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 10 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 QDPOrderProof (⇔, 71 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES

(0) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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:

HALF(s(s(x))) → HALF(x)
LASTBIT(s(s(x))) → LASTBIT(x)
CONV(x) → CONVITER(x, cons(0, nil))
CONVITER(x, l) → IF(zero(x), x, l)
CONVITER(x, l) → ZERO(x)
IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
IF(false, x, l) → HALF(x)
IF(false, x, l) → LASTBIT(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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 3 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LASTBIT(s(s(x))) → LASTBIT(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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:

LASTBIT(s(s(x))) → LASTBIT(x)

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

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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].

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(11) Obligation:

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

LASTBIT(s(s(x))) → LASTBIT(x)

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:

  • LASTBIT(s(s(x))) → LASTBIT(x)
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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:

HALF(s(s(x))) → HALF(x)

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

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(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].

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(18) Obligation:

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

HALF(s(s(x))) → HALF(x)

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:

  • HALF(s(s(x))) → HALF(x)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
CONVITER(x, l) → IF(zero(x), x, l)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, 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:

IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
CONVITER(x, l) → IF(zero(x), x, l)

The TRS R consists of the following rules:

zero(0) → true
zero(s(x)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))
conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, 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].

conv(x0)
conviter(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(25) Obligation:

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

IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
CONVITER(x, l) → IF(zero(x), x, l)

The TRS R consists of the following rules:

zero(0) → true
zero(s(x)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


IF(false, x, l) → CONVITER(half(x), cons(lastbit(x), l))
CONVITER(x, l) → IF(zero(x), x, l)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0   
POL(CONVITER(x1, x2)) = [1/4] + [1/2]x1 + x2   
POL(IF(x1, x2, x3)) = [1/4]x1 + [1/4]x2 + [1/2]x3   
POL(cons(x1, x2)) = 0   
POL(false) = [2]   
POL(half(x1)) = [1/4] + [1/4]x1   
POL(lastbit(x1)) = [4] + [1/4]x1   
POL(s(x1)) = [4] + [4]x1   
POL(true) = 0   
POL(zero(x1)) = [1/2]x1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
zero(0) → true
zero(s(x)) → false

(27) Obligation:

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

zero(0) → true
zero(s(x)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
lastbit(0)
lastbit(s(0))
lastbit(s(s(x0)))
zero(0)
zero(s(x0))

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