(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))
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:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))
The set Q consists of the following terms:
quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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:
QUOT(s(x), s(y), z) → QUOT(x, y, z)
PLUS(s(x), y) → PLUS(x, y)
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
QUOT(x, 0, s(z)) → PLUS(z, s(0))
The TRS R consists of the following rules:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))
The set Q consists of the following terms:
quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(x, y)
The TRS R consists of the following rules:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))
The set Q consists of the following terms:
quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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:
PLUS(s(x), y) → PLUS(x, y)
R is empty.
The set Q consists of the following terms:
quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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].
quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PLUS(s(x), y) → PLUS(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:
- PLUS(s(x), y) → PLUS(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
QUOT(s(x), s(y), z) → QUOT(x, y, z)
The TRS R consists of the following rules:
quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))
The set Q consists of the following terms:
quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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:
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
QUOT(s(x), s(y), z) → QUOT(x, y, z)
The TRS R consists of the following rules:
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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].
quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
quot(x0, 0, s(x1))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
QUOT(s(x), s(y), z) → QUOT(x, y, z)
The TRS R consists of the following rules:
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(19) 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(x, y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(
x1,
x2,
x3) =
x1
s(
x1) =
s(
x1)
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s_1=1
dummyConstant=1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
The TRS R consists of the following rules:
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
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(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(
x1,
x2,
x3) =
x2
0 =
0
plus(
x1,
x2) =
x2
s(
x1) =
s
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s=1
0=2
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
(22) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
The set Q consists of the following terms:
plus(0, x0)
plus(s(x0), x1)
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