(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)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
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(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, y)
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
MINUS(minus(x, y), z) → PLUS(y, z)
The TRS R consists of the following rules:
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(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
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:
PLUS(s(x), y) → PLUS(x, y)
The TRS R consists of the following rules:
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(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
PLUS(s(x), y) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
PLUS(
x0,
x1,
x2) =
PLUS(
x0,
x1)
Tags:
PLUS has argument tags [2,3,3] and root tag 0
Comparison: DMS
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
PLUS(
x1,
x2) =
PLUS
s(
x1) =
s(
x1)
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(7) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
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(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
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)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MINUS(minus(x, y), z) → MINUS(x, plus(y, z))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MINUS(
x0,
x1,
x2) =
MINUS(
x1)
Tags:
MINUS has argument tags [0,2,3] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
MINUS(
x1,
x2) =
x2
minus(
x1,
x2) =
minus(
x1,
x2)
plus(
x1,
x2) =
plus(
x2)
s(
x1) =
x1
0 =
0
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
(12) 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)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MINUS(s(x), s(y)) → MINUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MINUS(
x0,
x1,
x2) =
MINUS(
x2)
Tags:
MINUS has argument tags [0,2,2] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(14) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
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(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(16) TRUE
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The TRS R consists of the following rules:
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(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
QUOT(
x0,
x1,
x2) =
QUOT(
x0)
Tags:
QUOT has argument tags [0,3,2] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
QUOT(
x1,
x2) =
QUOT(
x1,
x2)
s(
x1) =
s(
x1)
minus(
x1,
x2) =
x1
0 =
0
plus(
x1,
x2) =
x1
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
minus(x, 0) → x
minus(minus(x, y), z) → minus(x, plus(y, z))
minus(s(x), s(y)) → minus(x, y)
(19) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
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(x, y), s(y)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
minus(minus(x, y), z) → minus(x, plus(y, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(21) TRUE