(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(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:
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
MINUS(minus(x, y), z) → PLUS(y, z)
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(x), y) → PLUS(y, x)
DIV(x, y) → QUOT(x, y, 0)
QUOT(s(x), s(y), z) → QUOT(minus(p(ack(0, x)), y), s(y), s(z))
QUOT(s(x), s(y), z) → MINUS(p(ack(0, x)), y)
QUOT(s(x), s(y), z) → P(ack(0, x))
QUOT(s(x), s(y), z) → ACK(0, x)
ACK(0, x) → PLUS(x, s(0))
ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), s(y)) → ACK(s(x), y)
The TRS R consists of the following rules:
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(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 4 SCCs with 6 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(y, x)
PLUS(s(x), y) → PLUS(x, s(y))
The TRS R consists of the following rules:
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(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:
PLUS(s(x), y) → PLUS(y, x)
PLUS(s(x), y) → PLUS(x, s(y))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
PLUS(s(x), y) → PLUS(y, x)
PLUS(s(x), y) → PLUS(x, s(y))
Used ordering: Knuth-Bendix order [KBO] with precedence:
s1 > PLUS2
and weight map:
s_1=1
PLUS_2=0
The variable weight is 1
(9) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)
The TRS R consists of the following rules:
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
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:
- ACK(s(x), 0) → ACK(x, s(0))
The graph contains the following edges 1 > 1
- ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
The graph contains the following edges 1 > 1
- ACK(s(x), s(y)) → ACK(s(x), y)
The graph contains the following edges 1 >= 1, 2 > 2
(14) YES
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
The TRS R consists of the following rules:
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) 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.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(s(x), s(y)) → MINUS(x, y)
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
The TRS R consists of the following rules:
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) 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
- MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
The graph contains the following edges 1 > 1
(19) YES
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y), z) → QUOT(minus(p(ack(0, x)), y), s(y), s(z))
The TRS R consists of the following rules:
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
QUOT(s(x), s(y), z) → QUOT(minus(p(ack(0, x)), y), s(y), s(z))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( QUOT(x1, ..., x3) ) = max{0, x1 - 1} |
| POL( minus(x1, x2) ) = x1 |
| POL( p(x1) ) = max{0, x1 - 2} |
| POL( ack(x1, x2) ) = x2 + 2 |
| POL( plus(x1, x2) ) = x1 + x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
p(s(x)) → x
p(0) → 0
minus(s(x), s(y)) → minus(x, y)
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
(22) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(y, x))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
p(0) → 0
div(x, y) → quot(x, y, 0)
quot(s(x), s(y), z) → quot(minus(p(ack(0, x)), y), s(y), s(z))
quot(0, s(y), z) → z
ack(0, x) → s(x)
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(24) YES