(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
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:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
(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:
DIVISION(x, y) → DIV(x, y, 0)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
DIV(x, y, z) → LT(x, y)
DIV(x, y, z) → INC(z)
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
IF(false, x, s(y), z) → MINUS(x, s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
LT(s(x), s(y)) → LT(x, y)
INC(s(x)) → INC(x)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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].
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
(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:
LT(s(x), s(y)) → LT(x, y)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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:
LT(s(x), s(y)) → LT(x, y)
R is empty.
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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].
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(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:
- LT(s(x), s(y)) → LT(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:
MINUS(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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].
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
(25) 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.
(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:
- MINUS(s(x), s(y)) → MINUS(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:
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
The TRS R consists of the following rules:
division(x, y) → div(x, y, 0)
div(x, y, z) → if(lt(x, y), x, y, inc(z))
if(true, x, y, z) → z
if(false, x, s(y), z) → div(minus(x, s(y)), s(y), z)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
The TRS R consists of the following rules:
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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].
division(x0, x1)
div(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x, y, z) → IF(lt(x, y), x, y, inc(z))
The TRS R consists of the following rules:
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(33) TransformationProof (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
DIV(
x,
y,
z) →
IF(
lt(
x,
y),
x,
y,
inc(
z)) at position [0] we obtained the following new rules [LPAR04]:
DIV(x0, 0, y2) → IF(false, x0, 0, inc(y2)) → DIV(x0, 0, y2) → IF(false, x0, 0, inc(y2))
DIV(0, s(x0), y2) → IF(true, 0, s(x0), inc(y2)) → DIV(0, s(x0), y2) → IF(true, 0, s(x0), inc(y2))
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2)) → DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
DIV(x0, 0, y2) → IF(false, x0, 0, inc(y2))
DIV(0, s(x0), y2) → IF(true, 0, s(x0), inc(y2))
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
The TRS R consists of the following rules:
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(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 1 SCC with 2 less nodes.
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
IF(false, x, s(y), z) → DIV(minus(x, s(y)), s(y), z)
The TRS R consists of the following rules:
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(37) TransformationProof (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
IF(
false,
x,
s(
y),
z) →
DIV(
minus(
x,
s(
y)),
s(
y),
z) at position [0] we obtained the following new rules [LPAR04]:
IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2) → IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
The TRS R consists of the following rules:
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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.
DIV(s(x0), s(x1), y2) → IF(lt(x0, x1), s(x0), s(x1), inc(y2))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( IF(x1, ..., x4) ) = max{0, x2 + 2x3 - 2} |
POL( lt(x1, x2) ) = 2x1 + 2x2 + 1 |
POL( DIV(x1, ..., x3) ) = x1 + 2x2 |
POL( minus(x1, x2) ) = x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
(40) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(x0), s(x1), y2) → DIV(minus(x0, x1), s(x1), y2)
The TRS R consists of the following rules:
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
inc(0) → s(0)
inc(s(x)) → s(inc(x))
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
inc(0)
inc(s(x0))
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