(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x
The TRS R 2 is
cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
The signature Sigma is {
cond1,
cond2}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(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:
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND1(s(x), y) → GR(s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, x, y) → COND1(p(x), y)
COND2(false, x, y) → P(x)
GR(s(x), s(y)) → GR(x, y)
NEQ(s(x), s(y)) → NEQ(x, y)
The TRS R consists of the following rules:
cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(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 3 SCCs with 2 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
NEQ(s(x), s(y)) → NEQ(x, y)
The TRS R consists of the following rules:
cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(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:
NEQ(s(x), s(y)) → NEQ(x, y)
R is empty.
The set Q consists of the following terms:
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(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].
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
NEQ(s(x), s(y)) → NEQ(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:
- NEQ(s(x), s(y)) → NEQ(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(13) TRUE
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GR(s(x), s(y)) → GR(x, y)
The TRS R consists of the following rules:
cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(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:
GR(s(x), s(y)) → GR(x, y)
R is empty.
The set Q consists of the following terms:
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(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].
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
GR(s(x), s(y)) → GR(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:
- GR(s(x), s(y)) → GR(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(20) TRUE
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)
The TRS R consists of the following rules:
cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x
The set Q consists of the following terms:
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(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:
COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(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].
cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(26) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
COND2(
true,
x,
y) →
COND1(
y,
y) we obtained the following new rules [LPAR04]:
COND2(true, s(z0), z1) → COND1(z1, z1)
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)
COND2(true, s(z0), z1) → COND1(z1, z1)
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(28) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
COND2(
false,
x,
y) →
COND1(
p(
x),
y) at position [0] we obtained the following new rules [LPAR04]:
COND2(false, 0, y1) → COND1(0, y1)
COND2(false, s(x0), y1) → COND1(x0, y1)
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, 0, y1) → COND1(0, y1)
COND2(false, s(x0), y1) → COND1(x0, y1)
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(30) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
The TRS R consists of the following rules:
p(0) → 0
p(s(x)) → x
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(32) 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.
(33) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
p(0)
p(s(x0))
We have to consider all minimal (P,Q,R)-chains.
(34) 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))
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(36) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
COND2(
true,
x,
y) →
COND1(
y,
y) we obtained the following new rules [LPAR04]:
COND2(true, s(z0), z1) → COND1(z1, z1)
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, s(x0), y1) → COND1(x0, y1)
COND2(true, s(z0), z1) → COND1(z1, z1)
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(38) 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
COND1(
s(
x),
y) →
COND2(
gr(
s(
x),
y),
s(
x),
y) the following chains were created:
- We consider the chain COND2(true, x, y) → COND1(y, y), COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) which results in the following constraint:
(1) (COND1(x3, x3)=COND1(s(x4), x5) ⇒ COND1(s(x4), x5)≥COND2(gr(s(x4), x5), s(x4), x5))
We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2) (COND1(s(x4), s(x4))≥COND2(gr(s(x4), s(x4)), s(x4), s(x4)))
- We consider the chain COND2(false, s(x0), y1) → COND1(x0, y1), COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) which results in the following constraint:
(3) (COND1(x6, x7)=COND1(s(x8), x9) ⇒ COND1(s(x8), x9)≥COND2(gr(s(x8), x9), s(x8), x9))
We simplified constraint (3) using rules (I), (II), (III) which results in the following new constraint:
(4) (COND1(s(x8), x7)≥COND2(gr(s(x8), x7), s(x8), x7))
For Pair
COND2(
true,
x,
y) →
COND1(
y,
y) the following chains were created:
- We consider the chain COND1(s(x), y) → COND2(gr(s(x), y), s(x), y), COND2(true, x, y) → COND1(y, y) which results in the following constraint:
(5) (COND2(gr(s(x10), x11), s(x10), x11)=COND2(true, x12, x13) ⇒ COND2(true, x12, x13)≥COND1(x13, x13))
We simplified constraint (5) using rules (I), (II), (III), (VII) which results in the following new constraint:
(6) (s(x10)=x26∧gr(x26, x11)=true ⇒ COND2(true, s(x10), x11)≥COND1(x11, x11))
We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on gr(x26, x11)=true which results in the following new constraints:
(7) (true=true∧s(x10)=s(x27) ⇒ COND2(true, s(x10), 0)≥COND1(0, 0))
(8) (gr(x29, x28)=true∧s(x10)=s(x29)∧(∀x30:gr(x29, x28)=true∧s(x30)=x29 ⇒ COND2(true, s(x30), x28)≥COND1(x28, x28)) ⇒ COND2(true, s(x10), s(x28))≥COND1(s(x28), s(x28)))
We simplified constraint (7) using rules (I), (II), (IV) which results in the following new constraint:
(9) (COND2(true, s(x10), 0)≥COND1(0, 0))
We simplified constraint (8) using rules (I), (II), (III), (IV) which results in the following new constraint:
(10) (gr(x29, x28)=true ⇒ COND2(true, s(x29), s(x28))≥COND1(s(x28), s(x28)))
We simplified constraint (10) using rule (V) (with possible (I) afterwards) using induction on gr(x29, x28)=true which results in the following new constraints:
(11) (true=true ⇒ COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))
(12) (gr(x34, x33)=true∧(gr(x34, x33)=true ⇒ COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33))) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))
We simplified constraint (11) using rules (I), (II) which results in the following new constraint:
(13) (COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))
We simplified constraint (12) using rule (VI) where we applied the induction hypothesis (gr(x34, x33)=true ⇒ COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33))) with σ = [ ] which results in the following new constraint:
(14) (COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33)) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))
For Pair
COND2(
false,
s(
x0),
y1) →
COND1(
x0,
y1) the following chains were created:
- We consider the chain COND1(s(x), y) → COND2(gr(s(x), y), s(x), y), COND2(false, s(x0), y1) → COND1(x0, y1) which results in the following constraint:
(15) (COND2(gr(s(x18), x19), s(x18), x19)=COND2(false, s(x20), x21) ⇒ COND2(false, s(x20), x21)≥COND1(x20, x21))
We simplified constraint (15) using rules (I), (II), (III), (VII) which results in the following new constraint:
(16) (s(x18)=x36∧gr(x36, x19)=false ⇒ COND2(false, s(x18), x19)≥COND1(x18, x19))
We simplified constraint (16) using rule (V) (with possible (I) afterwards) using induction on gr(x36, x19)=false which results in the following new constraints:
(17) (gr(x39, x38)=false∧s(x18)=s(x39)∧(∀x40:gr(x39, x38)=false∧s(x40)=x39 ⇒ COND2(false, s(x40), x38)≥COND1(x40, x38)) ⇒ COND2(false, s(x18), s(x38))≥COND1(x18, s(x38)))
(18) (false=false∧s(x18)=0 ⇒ COND2(false, s(x18), x41)≥COND1(x18, x41))
We simplified constraint (17) using rules (I), (II), (III), (IV) which results in the following new constraint:
(19) (gr(x39, x38)=false ⇒ COND2(false, s(x39), s(x38))≥COND1(x39, s(x38)))
We solved constraint (18) using rules (I), (II).We simplified constraint (19) using rule (V) (with possible (I) afterwards) using induction on gr(x39, x38)=false which results in the following new constraints:
(20) (gr(x44, x43)=false∧(gr(x44, x43)=false ⇒ COND2(false, s(x44), s(x43))≥COND1(x44, s(x43))) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))
(21) (false=false ⇒ COND2(false, s(0), s(x45))≥COND1(0, s(x45)))
We simplified constraint (20) using rule (VI) where we applied the induction hypothesis (gr(x44, x43)=false ⇒ COND2(false, s(x44), s(x43))≥COND1(x44, s(x43))) with σ = [ ] which results in the following new constraint:
(22) (COND2(false, s(x44), s(x43))≥COND1(x44, s(x43)) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))
We simplified constraint (21) using rules (I), (II) which results in the following new constraint:
(23) (COND2(false, s(0), s(x45))≥COND1(0, s(x45)))
To summarize, we get the following constraints P
≥ for the following pairs.
- COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
- (COND1(s(x4), s(x4))≥COND2(gr(s(x4), s(x4)), s(x4), s(x4)))
- (COND1(s(x8), x7)≥COND2(gr(s(x8), x7), s(x8), x7))
- COND2(true, x, y) → COND1(y, y)
- (COND2(true, s(x10), 0)≥COND1(0, 0))
- (COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))
- (COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33)) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))
- COND2(false, s(x0), y1) → COND1(x0, y1)
- (COND2(false, s(x44), s(x43))≥COND1(x44, s(x43)) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))
- (COND2(false, s(0), s(x45))≥COND1(0, s(x45)))
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(COND1(x1, x2)) = 1 + x1 + x2
POL(COND2(x1, x2, x3)) = 1 - x1 + x2 + x3
POL(c) = -1
POL(false) = 0
POL(gr(x1, x2)) = 0
POL(s(x1)) = 1 + x1
POL(true) = 1
The following pairs are in P
>:
COND2(false, s(x0), y1) → COND1(x0, y1)
The following pairs are in P
bound:
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)
The following rules are usable:
true → gr(s(x), 0)
gr(x, y) → gr(s(x), s(y))
false → gr(0, x)
(39) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(40) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
COND1(
s(
x),
y) →
COND2(
gr(
s(
x),
y),
s(
x),
y) we obtained the following new rules [LPAR04]:
COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))
(41) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND2(true, x, y) → COND1(y, y)
COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(42) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
COND1(
s(
x),
y) →
COND2(
gr(
s(
x),
y),
s(
x),
y) we obtained the following new rules [LPAR04]:
COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))
(43) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND2(true, x, y) → COND1(y, y)
COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(44) Rewriting (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
COND1(
s(
x0),
s(
x0)) →
COND2(
gr(
s(
x0),
s(
x0)),
s(
x0),
s(
x0)) at position [0] we obtained the following new rules [LPAR04]:
COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
(45) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND2(true, x, y) → COND1(y, y)
COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(46) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
COND2(
true,
x,
y) →
COND1(
y,
y) we obtained the following new rules [LPAR04]:
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))
(47) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(48) 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
COND1(
s(
x0),
s(
x0)) →
COND2(
gr(
x0,
x0),
s(
x0),
s(
x0)) the following chains were created:
- We consider the chain COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0)), COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0)) which results in the following constraint:
(1) (COND1(s(x1), s(x1))=COND1(s(x2), s(x2)) ⇒ COND1(s(x2), s(x2))≥COND2(gr(x2, x2), s(x2), s(x2)))
We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
(2) (COND1(s(x1), s(x1))≥COND2(gr(x1, x1), s(x1), s(x1)))
For Pair
COND2(
true,
s(
z0),
s(
z0)) →
COND1(
s(
z0),
s(
z0)) the following chains were created:
- We consider the chain COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0)), COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0)) which results in the following constraint:
(3) (COND2(gr(x3, x3), s(x3), s(x3))=COND2(true, s(x4), s(x4)) ⇒ COND2(true, s(x4), s(x4))≥COND1(s(x4), s(x4)))
We simplified constraint (3) using rules (I), (II), (III), (VII) which results in the following new constraint:
(4) (x3=x6∧gr(x3, x6)=true ⇒ COND2(true, s(x3), s(x3))≥COND1(s(x3), s(x3)))
We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on gr(x3, x6)=true which results in the following new constraints:
(5) (true=true∧s(x7)=0 ⇒ COND2(true, s(s(x7)), s(s(x7)))≥COND1(s(s(x7)), s(s(x7))))
(6) (gr(x9, x8)=true∧s(x9)=s(x8)∧(gr(x9, x8)=true∧x9=x8 ⇒ COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9))) ⇒ COND2(true, s(s(x9)), s(s(x9)))≥COND1(s(s(x9)), s(s(x9))))
We solved constraint (5) using rules (I), (II).We simplified constraint (6) using rules (I), (II) which results in the following new constraint:
(7) (gr(x9, x8)=true∧x9=x8∧(gr(x9, x8)=true∧x9=x8 ⇒ COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9))) ⇒ COND2(true, s(s(x9)), s(s(x9)))≥COND1(s(s(x9)), s(s(x9))))
We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (gr(x9, x8)=true∧x9=x8 ⇒ COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9))) with σ = [ ] which results in the following new constraint:
(8) (COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9)) ⇒ COND2(true, s(s(x9)), s(s(x9)))≥COND1(s(s(x9)), s(s(x9))))
To summarize, we get the following constraints P
≥ for the following pairs.
- COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
- (COND1(s(x1), s(x1))≥COND2(gr(x1, x1), s(x1), s(x1)))
- COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))
- (COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9)) ⇒ COND2(true, s(s(x9)), s(s(x9)))≥COND1(s(s(x9)), s(s(x9))))
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) = 1
POL(COND1(x1, x2)) = 1 - x1 - x2
POL(COND2(x1, x2, x3)) = -1 + x2 + x3
POL(c) = -1
POL(false) = 0
POL(gr(x1, x2)) = x2
POL(s(x1)) = 0
POL(true) = 1
The following pairs are in P
>:
COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))
The following pairs are in P
bound:
COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))
There are no usable rules
(49) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
gr(0, x) → false
The set Q consists of the following terms:
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(50) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(51) TRUE