YES Termination w.r.t. Q proof of AProVE_07_thiemann13.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 (⇔, 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 QDPOrderProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QReductionProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QReductionProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QReductionProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 QDPQMonotonicMRRProof (⇔, 10 ms)
        ↳43 QDP
        ↳44 QDPQMonotonicMRRProof (⇔, 13 ms)
        ↳45 QDP
        ↳46 TransformationProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 TransformationProof (⇔, 0 ms)
        ↳49 QDP
        ↳50 UsableRulesProof (⇔, 0 ms)
        ↳51 QDP
        ↳52 TransformationProof (⇒, 0 ms)
        ↳53 QDP
        ↳54 DependencyGraphProof (⇔, 0 ms)
        ↳55 QDP
        ↳56 TransformationProof (⇔, 0 ms)
        ↳57 QDP
        ↳58 TransformationProof (⇔, 0 ms)
        ↳59 QDP
        ↳60 TransformationProof (⇔, 0 ms)
        ↳61 QDP
        ↳62 UsableRulesProof (⇔, 0 ms)
        ↳63 QDP
        ↳64 QReductionProof (⇔, 0 ms)
        ↳65 QDP
        ↳66 TransformationProof (⇔, 0 ms)
        ↳67 QDP
        ↳68 DependencyGraphProof (⇔, 0 ms)
        ↳69 TRUE

(0) Obligation:

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

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))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), 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:

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))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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:

LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
LOG(x) → LOG2(x, 0)
LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
LOG2(x, y) → LE(x, 0)
LOG2(x, y) → LE(x, s(0))
LOG2(x, y) → INC(y)
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF2(false, x, y) → QUOT(x, s(s(0)))

The TRS R consists of the following rules:

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))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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 5 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

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))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(11) Obligation:

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

MINUS(s(x), s(y)) → MINUS(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:

  • MINUS(s(x), s(y)) → MINUS(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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

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))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(18) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))

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

(19) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  x1
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
0=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

(20) Obligation:

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

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))

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

(21) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(22) YES

(23) 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:

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))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(24) 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.

(25) 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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(26) 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(27) 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.

(28) 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

(29) YES

(30) 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:

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))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(31) 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.

(32) 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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(33) 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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(34) 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.

(35) 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

(36) YES

(37) Obligation:

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

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)

The TRS R consists of the following rules:

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))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(38) 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.

(39) Obligation:

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

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)

The TRS R consists of the following rules:

quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
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))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

(40) 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, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

(41) Obligation:

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

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)

The TRS R consists of the following rules:

quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
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))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(42) QDPQMonotonicMRRProof (EQUIVALENT transformation)

By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.

Strictly oriented rules of the TRS R:

minus(s(x), s(y)) → minus(x, y)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 2   
POL(IF(x1, x2, x3, x4)) = 1 + 2·x3 + x4   
POL(IF2(x1, x2, x3)) = 1 + 2·x2 + x3   
POL(LOG2(x1, x2)) = 1 + 2·x1 + x2   
POL(false) = 2   
POL(inc(x1)) = x1   
POL(le(x1, x2)) = 2 + x2   
POL(minus(x1, x2)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 1 + 2·x1   
POL(true) = 0   

(43) Obligation:

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

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)

The TRS R consists of the following rules:

quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → 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))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(44) QDPQMonotonicMRRProof (EQUIVALENT transformation)

By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.

Strictly oriented rules of the TRS R:

le(0, y) → true

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(IF(x1, x2, x3, x4)) = x1 + x2 + x4   
POL(IF2(x1, x2, x3)) = 1 + x1 + x3   
POL(LOG2(x1, x2)) = 2 + x2   
POL(false) = 1   
POL(inc(x1)) = x1   
POL(le(x1, x2)) = 1   
POL(minus(x1, x2)) = x1   
POL(quot(x1, x2)) = 1 + 2·x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

(45) Obligation:

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

LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))
IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)

The TRS R consists of the following rules:

quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(46) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y)) at position [0] we obtained the following new rules [LPAR04]:

LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1)) → LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1))

(47) Obligation:

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

IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1))

The TRS R consists of the following rules:

quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(48) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1)) at position [1] we obtained the following new rules [LPAR04]:

LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1)) → LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))

(49) Obligation:

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

IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))

The TRS R consists of the following rules:

quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(50) 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.

(51) Obligation:

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

IF(false, b, x, y) → IF2(b, x, y)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))

The TRS R consists of the following rules:

le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(52) TransformationProof (SOUND transformation)

By narrowing [LPAR04] the rule IF2(false, x, y) → LOG2(quot(x, s(s(0))), y) at position [0] we obtained the following new rules [LPAR04]:

IF2(false, 0, y1) → LOG2(0, y1) → IF2(false, 0, y1) → LOG2(0, y1)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1) → IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)

(53) Obligation:

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

IF(false, b, x, y) → IF2(b, x, y)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF2(false, 0, y1) → LOG2(0, y1)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)

The TRS R consists of the following rules:

le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(54) DependencyGraphProof (EQUIVALENT transformation)

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

(55) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(56) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, b, x, y) → IF2(b, x, y) we obtained the following new rules [LPAR04]:

IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1) → IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1)

(57) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1)

The TRS R consists of the following rules:

le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(58) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1) we obtained the following new rules [LPAR04]:

IF(false, false, s(x1), x2) → IF2(false, s(x1), x2) → IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

(59) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

The TRS R consists of the following rules:

le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(60) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1)) at position [1] we obtained the following new rules [LPAR04]:

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1)) → LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))

(61) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))

The TRS R consists of the following rules:

le(s(x), 0) → false
inc(0) → 0
inc(s(x)) → s(inc(x))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(0, y) → 0
minus(x, 0) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(62) 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.

(63) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))

The TRS R consists of the following rules:

inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(64) 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].

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

(65) Obligation:

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))

The TRS R consists of the following rules:

inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x

The set Q consists of the following terms:

inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(66) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1) at position [0,0] we obtained the following new rules [LPAR04]:

IF2(false, s(0), y1) → LOG2(s(quot(0, s(s(0)))), y1) → IF2(false, s(0), y1) → LOG2(s(quot(0, s(s(0)))), y1)

(67) Obligation:

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

IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF2(false, s(0), y1) → LOG2(s(quot(0, s(s(0)))), y1)

The TRS R consists of the following rules:

inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x

The set Q consists of the following terms:

inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))

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

(68) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(69) TRUE