(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), 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:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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:
LE(s(x), s(y)) → LE(x, y)
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
TIMES(s(x), y) → TIMES(x, y)
LOG(s(x), s(s(b))) → LOOP(s(x), s(s(b)), s(0), 0)
LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
LOOP(x, s(s(b)), s(y), z) → LE(x, s(y))
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
IF(false, x, b, y, z) → TIMES(b, y)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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 4 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:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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].
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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:
TIMES(s(x), y) → TIMES(x, y)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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:
TIMES(s(x), y) → TIMES(x, y)
R is empty.
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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].
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TIMES(s(x), y) → TIMES(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:
- TIMES(s(x), y) → TIMES(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:
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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:
LE(s(x), s(y)) → LE(x, y)
R is empty.
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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].
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
(25) 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.
(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:
- LE(s(x), s(y)) → LE(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(27) YES
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
The TRS R consists of the following rules:
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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:
LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, 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].
log(x0, 0)
log(x0, s(0))
log(0, s(s(x0)))
log(s(x0), s(s(x1)))
loop(x0, s(s(x1)), s(x2), x3)
if(true, x0, x1, x2, x3)
if(false, x0, x1, x2, x3)
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(x, s(s(b)), s(y), z) → IF(le(x, s(y)), x, s(s(b)), s(y), z)
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(33) TransformationProof (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
LOOP(
x,
s(
s(
b)),
s(
y),
z) →
IF(
le(
x,
s(
y)),
x,
s(
s(
b)),
s(
y),
z) at position [0] we obtained the following new rules [LPAR04]:
LOOP(0, s(s(y1)), s(y2), y3) → IF(true, 0, s(s(y1)), s(y2), y3) → LOOP(0, s(s(y1)), s(y2), y3) → IF(true, 0, s(s(y1)), s(y2), y3)
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3) → LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
LOOP(0, s(s(y1)), s(y2), y3) → IF(true, 0, s(s(y1)), s(y2), y3)
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(35) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, x, b, y, z) → LOOP(x, b, times(b, y), s(z))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(37) TransformationProof (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
IF(
false,
x,
b,
y,
z) →
LOOP(
x,
b,
times(
b,
y),
s(
z)) at position [2] we obtained the following new rules [LPAR04]:
IF(false, y0, 0, x0, y3) → LOOP(y0, 0, 0, s(y3)) → IF(false, y0, 0, x0, y3) → LOOP(y0, 0, 0, s(y3))
IF(false, y0, s(x0), x1, y3) → LOOP(y0, s(x0), plus(x1, times(x0, x1)), s(y3)) → IF(false, y0, s(x0), x1, y3) → LOOP(y0, s(x0), plus(x1, times(x0, x1)), s(y3))
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, y0, 0, x0, y3) → LOOP(y0, 0, 0, s(y3))
IF(false, y0, s(x0), x1, y3) → LOOP(y0, s(x0), plus(x1, times(x0, x1)), s(y3))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(39) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(40) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, y0, s(x0), x1, y3) → LOOP(y0, s(x0), plus(x1, times(x0, x1)), s(y3))
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(41) TransformationProof (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
IF(
false,
y0,
s(
x0),
x1,
y3) →
LOOP(
y0,
s(
x0),
plus(
x1,
times(
x0,
x1)),
s(
y3)) we obtained the following new rules [LPAR04]:
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), plus(s(z2), times(s(z1), s(z2))), s(z3)) → IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), plus(s(z2), times(s(z1), s(z2))), s(z3))
(42) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), plus(s(z2), times(s(z1), s(z2))), s(z3))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(43) TransformationProof (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
IF(
false,
s(
z0),
s(
s(
z1)),
s(
z2),
z3) →
LOOP(
s(
z0),
s(
s(
z1)),
plus(
s(
z2),
times(
s(
z1),
s(
z2))),
s(
z3)) at position [2] we obtained the following new rules [LPAR04]:
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, times(s(z1), s(z2)))), s(z3)) → IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, times(s(z1), s(z2)))), s(z3))
(44) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, times(s(z1), s(z2)))), s(z3))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(45) TransformationProof (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
IF(
false,
s(
z0),
s(
s(
z1)),
s(
z2),
z3) →
LOOP(
s(
z0),
s(
s(
z1)),
s(
plus(
z2,
times(
s(
z1),
s(
z2)))),
s(
z3)) at position [2,0,1] we obtained the following new rules [LPAR04]:
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, plus(s(z2), times(z1, s(z2))))), s(z3)) → IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, plus(s(z2), times(z1, s(z2))))), s(z3))
(46) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, plus(s(z2), times(z1, s(z2))))), s(z3))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(47) TransformationProof (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
IF(
false,
s(
z0),
s(
s(
z1)),
s(
z2),
z3) →
LOOP(
s(
z0),
s(
s(
z1)),
s(
plus(
z2,
plus(
s(
z2),
times(
z1,
s(
z2))))),
s(
z3)) at position [2,0,1] we obtained the following new rules [LPAR04]:
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(z3)) → IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(z3))
(48) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(s(x0), s(s(y1)), s(x1), y3) → IF(le(x0, x1), s(x0), s(s(y1)), s(x1), y3)
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(z3))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(49) TransformationProof (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
LOOP(
s(
x0),
s(
s(
y1)),
s(
x1),
y3) →
IF(
le(
x0,
x1),
s(
x0),
s(
s(
y1)),
s(
x1),
y3) we obtained the following new rules [LPAR04]:
LOOP(s(z0), s(s(z1)), s(y_2), s(z3)) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(z3)) → LOOP(s(z0), s(s(z1)), s(y_2), s(z3)) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(z3))
(50) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(z0), s(s(z1)), s(z2), z3) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(z3))
LOOP(s(z0), s(s(z1)), s(y_2), s(z3)) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(z3))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(51) TransformationProof (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
IF(
false,
s(
z0),
s(
s(
z1)),
s(
z2),
z3) →
LOOP(
s(
z0),
s(
s(
z1)),
s(
plus(
z2,
s(
plus(
z2,
times(
z1,
s(
z2)))))),
s(
z3)) we obtained the following new rules [LPAR04]:
IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3))) → IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))
(52) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(s(z0), s(s(z1)), s(y_2), s(z3)) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(z3))
IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(53) TransformationProof (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
LOOP(
s(
z0),
s(
s(
z1)),
s(
y_2),
s(
z3)) →
IF(
le(
z0,
y_2),
s(
z0),
s(
s(
z1)),
s(
y_2),
s(
z3)) we obtained the following new rules [LPAR04]:
LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3))) → LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3)))
(54) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))
LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3)))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(55) 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
IF(
false,
s(
z0),
s(
s(
z1)),
s(
z2),
s(
z3)) →
LOOP(
s(
z0),
s(
s(
z1)),
s(
plus(
z2,
s(
plus(
z2,
times(
z1,
s(
z2)))))),
s(
s(
z3))) the following chains were created:
- We consider the chain LOOP(s(x4), s(s(x5)), s(x6), s(s(x7))) → IF(le(x4, x6), s(x4), s(s(x5)), s(x6), s(s(x7))), IF(false, s(x8), s(s(x9)), s(x10), s(x11)) → LOOP(s(x8), s(s(x9)), s(plus(x10, s(plus(x10, times(x9, s(x10)))))), s(s(x11))) which results in the following constraint:
| (1) (IF(le(x4, x6), s(x4), s(s(x5)), s(x6), s(s(x7)))=IF(false, s(x8), s(s(x9)), s(x10), s(x11)) ⇒ IF(false, s(x8), s(s(x9)), s(x10), s(x11))≥LOOP(s(x8), s(s(x9)), s(plus(x10, s(plus(x10, times(x9, s(x10)))))), s(s(x11)))) |
We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
| (2) (le(x4, x6)=false ⇒ IF(false, s(x4), s(s(x5)), s(x6), s(s(x7)))≥LOOP(s(x4), s(s(x5)), s(plus(x6, s(plus(x6, times(x5, s(x6)))))), s(s(s(x7))))) |
We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x4, x6)=false which results in the following new constraints:
| (3) (le(x26, x25)=false∧(∀x27,x28:le(x26, x25)=false ⇒ IF(false, s(x26), s(s(x27)), s(x25), s(s(x28)))≥LOOP(s(x26), s(s(x27)), s(plus(x25, s(plus(x25, times(x27, s(x25)))))), s(s(s(x28))))) ⇒ IF(false, s(s(x26)), s(s(x5)), s(s(x25)), s(s(x7)))≥LOOP(s(s(x26)), s(s(x5)), s(plus(s(x25), s(plus(s(x25), times(x5, s(s(x25))))))), s(s(s(x7))))) |
| (4) (false=false ⇒ IF(false, s(s(x29)), s(s(x5)), s(0), s(s(x7)))≥LOOP(s(s(x29)), s(s(x5)), s(plus(0, s(plus(0, times(x5, s(0)))))), s(s(s(x7))))) |
We simplified constraint (3) using rule (VI) where we applied the induction hypothesis (∀x27,x28:le(x26, x25)=false ⇒ IF(false, s(x26), s(s(x27)), s(x25), s(s(x28)))≥LOOP(s(x26), s(s(x27)), s(plus(x25, s(plus(x25, times(x27, s(x25)))))), s(s(s(x28))))) with σ = [x27 / x5, x28 / x7] which results in the following new constraint:
| (5) (IF(false, s(x26), s(s(x5)), s(x25), s(s(x7)))≥LOOP(s(x26), s(s(x5)), s(plus(x25, s(plus(x25, times(x5, s(x25)))))), s(s(s(x7)))) ⇒ IF(false, s(s(x26)), s(s(x5)), s(s(x25)), s(s(x7)))≥LOOP(s(s(x26)), s(s(x5)), s(plus(s(x25), s(plus(s(x25), times(x5, s(s(x25))))))), s(s(s(x7))))) |
We simplified constraint (4) using rules (I), (II) which results in the following new constraint:
| (6) (IF(false, s(s(x29)), s(s(x5)), s(0), s(s(x7)))≥LOOP(s(s(x29)), s(s(x5)), s(plus(0, s(plus(0, times(x5, s(0)))))), s(s(s(x7))))) |
For Pair
LOOP(
s(
z0),
s(
s(
z1)),
s(
y_2),
s(
s(
z3))) →
IF(
le(
z0,
y_2),
s(
z0),
s(
s(
z1)),
s(
y_2),
s(
s(
z3))) the following chains were created:
- We consider the chain IF(false, s(x12), s(s(x13)), s(x14), s(x15)) → LOOP(s(x12), s(s(x13)), s(plus(x14, s(plus(x14, times(x13, s(x14)))))), s(s(x15))), LOOP(s(x16), s(s(x17)), s(x18), s(s(x19))) → IF(le(x16, x18), s(x16), s(s(x17)), s(x18), s(s(x19))) which results in the following constraint:
| (1) (LOOP(s(x12), s(s(x13)), s(plus(x14, s(plus(x14, times(x13, s(x14)))))), s(s(x15)))=LOOP(s(x16), s(s(x17)), s(x18), s(s(x19))) ⇒ LOOP(s(x16), s(s(x17)), s(x18), s(s(x19)))≥IF(le(x16, x18), s(x16), s(s(x17)), s(x18), s(s(x19)))) |
We simplified constraint (1) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:
| (2) (LOOP(s(x12), s(s(x13)), s(x18), s(s(x15)))≥IF(le(x12, x18), s(x12), s(s(x13)), s(x18), s(s(x15)))) |
To summarize, we get the following constraints P
≥ for the following pairs.
- IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))
- (IF(false, s(x26), s(s(x5)), s(x25), s(s(x7)))≥LOOP(s(x26), s(s(x5)), s(plus(x25, s(plus(x25, times(x5, s(x25)))))), s(s(s(x7)))) ⇒ IF(false, s(s(x26)), s(s(x5)), s(s(x25)), s(s(x7)))≥LOOP(s(s(x26)), s(s(x5)), s(plus(s(x25), s(plus(s(x25), times(x5, s(s(x25))))))), s(s(s(x7)))))
- (IF(false, s(s(x29)), s(s(x5)), s(0), s(s(x7)))≥LOOP(s(s(x29)), s(s(x5)), s(plus(0, s(plus(0, times(x5, s(0)))))), s(s(s(x7)))))
- LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3)))
- (LOOP(s(x12), s(s(x13)), s(x18), s(s(x15)))≥IF(le(x12, x18), s(x12), s(s(x13)), s(x18), s(s(x15))))
The constraints for P
> respective P
bound 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 P
bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:
POL(0) = 0
POL(IF(x1, x2, x3, x4, x5)) = 1 + x2 - x4
POL(LOOP(x1, x2, x3, x4)) = 1 + x1 - x3
POL(c) = -1
POL(false) = 1
POL(le(x1, x2)) = 0
POL(plus(x1, x2)) = x1 + x2
POL(s(x1)) = 1 + x1
POL(times(x1, x2)) = 0
POL(true) = 1
The following pairs are in P
>:
IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))
The following pairs are in P
bound:
IF(false, s(z0), s(s(z1)), s(z2), s(z3)) → LOOP(s(z0), s(s(z1)), s(plus(z2, s(plus(z2, times(z1, s(z2)))))), s(s(z3)))
The following rules are usable:
0 → times(0, y)
plus(y, times(x, y)) → times(s(x), y)
y → plus(0, y)
s(plus(x, y)) → plus(s(x), y)
(56) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LOOP(s(z0), s(s(z1)), s(y_2), s(s(z3))) → IF(le(z0, y_2), s(z0), s(s(z1)), s(y_2), s(s(z3)))
The TRS R consists of the following rules:
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
The set Q consists of the following terms:
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(57) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(58) TRUE