YES Termination w.r.t. Q proof of AProVE_07_thiemann12.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 7 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 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QReductionProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QDPOrderProof (⇔, 52 ms)
        ↳34 QDP
        ↳35 DependencyGraphProof (⇔, 0 ms)
        ↳36 TRUE

(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))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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)
LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
LOG2(x, y) → LE(x, s(0))
LOG2(x, y) → INC(y)
IF(false, x, y) → LOG2(half(x), y)
IF(false, x, y) → HALF(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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:

INC(s(x)) → INC(x)

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

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

(11) Obligation:

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

INC(s(x)) → INC(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:

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

(13) YES

(14) 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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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:

LE(s(x), s(y)) → LE(x, y)

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

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

(18) 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.

(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:

  • LE(s(x), s(y)) → LE(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:

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))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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:

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)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(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].

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

(25) 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.

(26) 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

(27) YES

(28) Obligation:

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

IF(false, x, y) → LOG2(half(x), y)
LOG2(x, y) → IF(le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, s(0)), x, inc(y))
if(true, x, s(y)) → y
if(false, x, y) → log2(half(x), y)

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

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

(29) 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.

(30) Obligation:

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

IF(false, x, y) → LOG2(half(x), y)
LOG2(x, y) → IF(le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

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

(31) 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].

log(x0)
log2(x0, x1)
if(true, x0, s(x1))
if(false, x0, x1)

(32) Obligation:

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

IF(false, x, y) → LOG2(half(x), y)
LOG2(x, y) → IF(le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))

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

(33) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


LOG2(x, y) → IF(le(x, s(0)), x, inc(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0   
POL(IF(x1, x2, x3)) = [1/4]x1 + [2]x2 + [1/4]x3   
POL(LOG2(x1, x2)) = [1/2] + [4]x1 + [1/4]x2   
POL(false) = [2]   
POL(half(x1)) = [1/2]x1   
POL(inc(x1)) = [1/4] + x1   
POL(le(x1, x2)) = [1] + [4]x1 + [1/4]x2   
POL(s(x1)) = [1/4] + x1   
POL(true) = 0   
The value of delta used in the strict ordering is 11/64.
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))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
le(s(x), 0) → false

(34) Obligation:

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

IF(false, x, y) → LOG2(half(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))

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

(35) DependencyGraphProof (EQUIVALENT transformation)

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

(36) TRUE