YES Termination w.r.t. Q proof of AProVE_07_thiemann16.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 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 (⇔, 84 ms)
        ↳34 QDP
        ↳35 UsableRulesProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 QReductionProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 QDPOrderProof (⇔, 29 ms)
        ↳40 QDP
        ↳41 DependencyGraphProof (⇔, 0 ms)
        ↳42 TRUE

(0) Obligation:

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

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

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:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

(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:

CHECK(s(s(s(x)))) → CHECK(s(x))
HALF(s(s(x))) → HALF(x)
PLUS(s(x), y) → PLUS(x, y)
TIMES(x, y) → TIMESITER(x, y, 0)
TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
TIMESITER(x, y, z) → CHECK(x)
TIMESITER(x, y, z) → PLUS(z, y)
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)
IF(odd, x, y, z, u) → P(x)
IF(even, x, y, z, u) → PLUS(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(z))
IF(even, x, y, z, u) → HALF(x)
IF(even, x, y, z, u) → HALF(z)
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(s(z)))
IF(even, x, y, z, u) → HALF(s(z))

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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(s(x), y) → PLUS(x, y)

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

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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].

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

(11) Obligation:

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

PLUS(s(x), y) → PLUS(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:

  • PLUS(s(x), y) → PLUS(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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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].

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

(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:

CHECK(s(s(s(x)))) → CHECK(s(x))

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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:

CHECK(s(s(s(x)))) → CHECK(s(x))

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

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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].

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

(25) Obligation:

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

CHECK(s(s(s(x)))) → CHECK(s(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:

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

(27) YES

(28) Obligation:

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

TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(z))
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(s(z)))

The TRS R consists of the following rules:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → if(check(x), x, y, z, plus(z, y))
p(s(x)) → x
p(0) → 0
if(zero, x, y, z, u) → z
if(odd, x, y, z, u) → timesIter(p(x), y, u)
if(even, x, y, z, u) → plus(timesIter(half(x), y, half(z)), timesIter(half(x), y, half(s(z))))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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:

TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(z))
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(s(z)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
p(s(x)) → x
p(0) → 0
check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
times(x0, x1)
timesIter(x0, x1, x2)
p(s(x0))
p(0)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

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].

times(x0, x1)
timesIter(x0, x1, x2)
if(zero, x0, x1, x2, x3)
if(odd, x0, x1, x2, x3)
if(even, x0, x1, x2, x3)

(32) Obligation:

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

TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(z))
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(s(z)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
p(s(x)) → x
p(0) → 0
check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
p(s(x0))
p(0)

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.


IF(even, x, y, z, u) → TIMESITER(half(x), y, half(z))
IF(even, x, y, z, u) → TIMESITER(half(x), y, half(s(z)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0   
POL(IF(x1, x2, x3, x4, x5)) = [4] + [1/4]x1 + [2]x2 + x3   
POL(TIMESITER(x1, x2, x3)) = [4] + [4]x1 + x2   
POL(check(x1)) = [2]x1   
POL(even) = [1/2]   
POL(half(x1)) = [1/2]x1   
POL(odd) = 0   
POL(p(x1)) = [1/2]x1   
POL(plus(x1, x2)) = [4]x1 + x2   
POL(s(x1)) = [4] + [2]x1   
POL(zero) = 0   
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:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
p(s(x)) → x
p(0) → 0
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

(34) Obligation:

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

TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
p(s(x)) → x
p(0) → 0
check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
p(s(x0))
p(0)

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

(35) 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.

(36) Obligation:

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

TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
half(0)
half(s(0))
half(s(s(x0)))
plus(0, x0)
plus(s(x0), x1)
p(s(x0))
p(0)

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

(37) 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)))

(38) Obligation:

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

TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))
IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
plus(0, x0)
plus(s(x0), x1)
p(s(x0))
p(0)

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

(39) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


IF(odd, x, y, z, u) → TIMESITER(p(x), y, u)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = [4]   
POL(IF(x1, x2, x3, x4, x5)) = [2]x1 + [1/4]x2   
POL(TIMESITER(x1, x2, x3)) = [2] + x1   
POL(check(x1)) = [1] + [1/4]x1   
POL(even) = [1]   
POL(odd) = [4]   
POL(p(x1)) = [4] + [1/4]x1   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = [4] + [4]x1   
POL(zero) = [1/4]   
The value of delta used in the strict ordering is 2.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
p(s(x)) → x
p(0) → 0

(40) Obligation:

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

TIMESITER(x, y, z) → IF(check(x), x, y, z, plus(z, y))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
check(0) → zero
check(s(0)) → odd
check(s(s(0))) → even
check(s(s(s(x)))) → check(s(x))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

The set Q consists of the following terms:

check(0)
check(s(0))
check(s(s(0)))
check(s(s(s(x0))))
plus(0, x0)
plus(s(x0), x1)
p(s(x0))
p(0)

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

(41) DependencyGraphProof (EQUIVALENT transformation)

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

(42) TRUE