(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x))) → P(x)
LE(p(s(x)), x) → LE(x, x)
LE(s(x), s(y)) → LE(x, y)
MINUS(x, y) → IF(le(x, y), x, y)
MINUS(x, y) → LE(x, y)
IF(false, x, y) → MINUS(p(x), y)
IF(false, x, y) → P(x)
The TRS R consists of the following rules:
p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
LE(p(s(x)), x) → LE(x, x)
The TRS R consists of the following rules:
p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
LE(p(s(x)), x) → LE(x, x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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
- LE(p(s(x)), x) → LE(x, x)
The graph contains the following edges 1 > 1, 2 >= 1, 1 > 2, 2 >= 2
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x))) → P(x)
The TRS R consists of the following rules:
p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
P(p(s(x))) → P(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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:
- P(p(s(x))) → P(x)
The graph contains the following edges 1 > 1
(14) TRUE
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), x, y)
The TRS R consists of the following rules:
p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), x, y)
The TRS R consists of the following rules:
p(0) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
(18) 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.
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), 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)
p(0) → s(s(0))
p(s(x)) → x
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
We have to consider all minimal (P,Q,R)-chains.
(20) 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].
minus(x0, x1)
if(true, x0, x1)
if(false, x0, x1)
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, y) → MINUS(p(x), y)
MINUS(x, y) → IF(le(x, y), 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)
p(0) → s(s(0))
p(s(x)) → x
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(22) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
MINUS(
x,
y) →
IF(
le(
x,
y),
x,
y) at position [0] we obtained the following new rules [LPAR04]:
MINUS(0, x0) → IF(true, 0, x0)
MINUS(s(x0), 0) → IF(false, s(x0), 0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, x, y) → MINUS(p(x), y)
MINUS(0, x0) → IF(true, 0, x0)
MINUS(s(x0), 0) → IF(false, s(x0), 0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
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)
p(0) → s(s(0))
p(s(x)) → x
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(24) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, x, y) → MINUS(p(x), y)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
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)
p(0) → s(s(0))
p(s(x)) → x
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(26) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
IF(
false,
x,
y) →
MINUS(
p(
x),
y) at position [0] we obtained the following new rules [LPAR04]:
IF(false, 0, y1) → MINUS(s(s(0)), y1)
IF(false, s(x0), y1) → MINUS(x0, y1)
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x0), 0) → IF(false, s(x0), 0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
IF(false, 0, y1) → MINUS(s(s(0)), y1)
IF(false, s(x0), y1) → MINUS(x0, y1)
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)
p(0) → s(s(0))
p(s(x)) → x
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(28) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(x0), y1) → MINUS(x0, y1)
MINUS(s(x0), 0) → IF(false, s(x0), 0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
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)
p(0) → s(s(0))
p(s(x)) → x
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(30) 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.
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(x0), y1) → MINUS(x0, y1)
MINUS(s(x0), 0) → IF(false, s(x0), 0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
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)
The set Q consists of the following terms:
p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(32) 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].
p(0)
p(s(x0))
(33) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(x0), y1) → MINUS(x0, y1)
MINUS(s(x0), 0) → IF(false, s(x0), 0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
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)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(34) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
IF(
false,
s(
x0),
y1) →
MINUS(
x0,
y1) we obtained the following new rules [LPAR04]:
IF(false, s(z0), 0) → MINUS(z0, 0)
IF(false, s(z0), s(z1)) → MINUS(z0, s(z1))
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x0), 0) → IF(false, s(x0), 0)
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
IF(false, s(z0), 0) → MINUS(z0, 0)
IF(false, s(z0), s(z1)) → MINUS(z0, s(z1))
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)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(36) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(37) Complex Obligation (AND)
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(z0), s(z1)) → MINUS(z0, s(z1))
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
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)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(39) ForwardInstantiation (EQUIVALENT transformation)
By forward instantiating [JAR06] the rule
IF(
false,
s(
z0),
s(
z1)) →
MINUS(
z0,
s(
z1)) we obtained the following new rules [LPAR04]:
IF(false, s(s(y_0)), s(x1)) → MINUS(s(y_0), s(x1))
(40) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x0), s(x1)) → IF(le(x0, x1), s(x0), s(x1))
IF(false, s(s(y_0)), s(x1)) → MINUS(s(y_0), s(x1))
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)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(41) ForwardInstantiation (EQUIVALENT transformation)
By forward instantiating [JAR06] the rule
MINUS(
s(
x0),
s(
x1)) →
IF(
le(
x0,
x1),
s(
x0),
s(
x1)) we obtained the following new rules [LPAR04]:
MINUS(s(s(y_1)), s(x1)) → IF(le(s(y_1), x1), s(s(y_1)), s(x1))
(42) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(s(y_0)), s(x1)) → MINUS(s(y_0), s(x1))
MINUS(s(s(y_1)), s(x1)) → IF(le(s(y_1), x1), s(s(y_1)), s(x1))
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)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(43) 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(s(y_1)), s(x1)) → IF(le(s(y_1), x1), s(s(y_1)), s(x1))
The graph contains the following edges 1 >= 2, 2 >= 3
- IF(false, s(s(y_0)), s(x1)) → MINUS(s(y_0), s(x1))
The graph contains the following edges 2 > 1, 3 >= 2
(44) TRUE
(45) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(z0), 0) → MINUS(z0, 0)
MINUS(s(x0), 0) → IF(false, s(x0), 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)
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(46) 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.
(47) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(z0), 0) → MINUS(z0, 0)
MINUS(s(x0), 0) → IF(false, s(x0), 0)
R is empty.
The set Q consists of the following terms:
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(48) 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))
(49) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(z0), 0) → MINUS(z0, 0)
MINUS(s(x0), 0) → IF(false, s(x0), 0)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(50) ForwardInstantiation (EQUIVALENT transformation)
By forward instantiating [JAR06] the rule
IF(
false,
s(
z0),
0) →
MINUS(
z0,
0) we obtained the following new rules [LPAR04]:
IF(false, s(s(y_0)), 0) → MINUS(s(y_0), 0)
(51) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x0), 0) → IF(false, s(x0), 0)
IF(false, s(s(y_0)), 0) → MINUS(s(y_0), 0)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(52) ForwardInstantiation (EQUIVALENT transformation)
By forward instantiating [JAR06] the rule
MINUS(
s(
x0),
0) →
IF(
false,
s(
x0),
0) we obtained the following new rules [LPAR04]:
MINUS(s(s(y_0)), 0) → IF(false, s(s(y_0)), 0)
(53) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(false, s(s(y_0)), 0) → MINUS(s(y_0), 0)
MINUS(s(s(y_0)), 0) → IF(false, s(s(y_0)), 0)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(54) 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(s(y_0)), 0) → IF(false, s(s(y_0)), 0)
The graph contains the following edges 1 >= 2, 2 >= 3
- IF(false, s(s(y_0)), 0) → MINUS(s(y_0), 0)
The graph contains the following edges 2 > 1, 3 >= 2
(55) TRUE