(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The TRS R 2 is
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
The signature Sigma is {
le,
true,
false}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
(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:
MINUS(s(x), s(y)) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
QUOT(s(x), s(y)) → MINUS(s(x), s(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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 1 less node.
(6) Complex Obligation (AND)
(7) 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:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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:
LE(s(x), s(y)) → LE(x, y)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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].
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
(11) 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.
(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:
- LE(s(x), s(y)) → LE(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:
MINUS(s(x), s(y)) → MINUS(x, y)
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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:
MINUS(s(x), s(y)) → MINUS(x, y)
R is empty.
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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].
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
(18) 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.
(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:
- MINUS(s(x), s(y)) → MINUS(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:
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
The TRS R consists of the following rules:
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))
The set Q consists of the following terms:
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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:
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
The TRS R consists of the following rules:
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))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))
The TRS R consists of the following rules:
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))
We have to consider all minimal (P,Q,R)-chains.
(26) Induction-Processor (SOUND transformation)
This DP could be deleted by the Induction-Processor:QUOT(
s(
x),
s(
y)) →
QUOT(
minus(
s(
x),
s(
y)),
s(
y))
This order was computed:Polynomial interpretation [POLO]:
POL(0) = 0
POL(QUOT(x1, x2)) = x1
POL(minus(x1, x2)) = x1
POL(s(x1)) = 1 + x1
At least one of these decreasing rules is always used after the deleted DP:minus(
s(
x),
s(
y)) →
minus(
x,
y)
The following formula is valid:x:sort[a0],y:sort[a0].
minus'(
s(
x ),
s(
y ))=
trueThe transformed set:minus'(
s(
x),
s(
y)) →
trueminus'(
x1,
0) →
falseminus'(
0,
s(
x0)) →
falseminus'(
s(
x2),
0) →
falseminus(
s(
x),
s(
y)) →
minus(
x,
y)
minus(
x1,
0) →
x1minus(
0,
s(
x0)) →
0minus(
s(
x2),
0) →
0equal_bool(
true,
false) →
falseequal_bool(
false,
true) →
falseequal_bool(
true,
true) →
trueequal_bool(
false,
false) →
trueand(
true,
x) →
xand(
false,
x) →
falseor(
true,
x) →
trueor(
false,
x) →
xnot(
false) →
truenot(
true) →
falseisa_true(
true) →
trueisa_true(
false) →
falseisa_false(
true) →
falseisa_false(
false) →
trueequal_sort[a0](
0,
0) →
trueequal_sort[a0](
0,
s(
x0)) →
falseequal_sort[a0](
s(
x0),
0) →
falseequal_sort[a0](
s(
x0),
s(
x1)) →
equal_sort[a0](
x0,
x1)
equal_sort[a11](
witness_sort[a11],
witness_sort[a11]) →
true
(27) Complex Obligation (AND)
(28) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
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))
We have to consider all minimal (P,Q,R)-chains.
(29) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(30) TRUE
(31) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus'(s(x), s(y)) → true
minus'(x1, 0) → false
minus'(0, s(x0)) → false
minus'(s(x2), 0) → false
minus(s(x), s(y)) → minus(x, y)
minus(x1, 0) → x1
minus(0, s(x0)) → 0
minus(s(x2), 0) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a11](witness_sort[a11], witness_sort[a11]) → true
Q is empty.
(32) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:
minus'2 > true
minus'2 > false
equalbool2 > true
not1 > true
not1 > false
isafalse1 > true
isafalse1 > false
equalsort[a0]2 > true
equalsort[a0]2 > false
equalsort[a11]2 > true
Status:
trivial
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
minus'(s(x), s(y)) → true
minus'(x1, 0) → false
minus'(0, s(x0)) → false
minus'(s(x2), 0) → false
minus(s(x), s(y)) → minus(x, y)
minus(x1, 0) → x1
minus(0, s(x0)) → 0
minus(s(x2), 0) → 0
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a11](witness_sort[a11], witness_sort[a11]) → true
(33) Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.
(34) RisEmptyProof (EQUIVALENT transformation)
The TRS R is empty. Hence, termination is trivially proven.
(35) TRUE
(36) RisEmptyProof (EQUIVALENT transformation)
The TRS R is empty. Hence, termination is trivially proven.
(37) TRUE
(38) RisEmptyProof (EQUIVALENT transformation)
The TRS R is empty. Hence, termination is trivially proven.
(39) TRUE