(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(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:
PLUS(s(X), Y) → PLUS(X, Y)
MIN(s(X), s(Y)) → MIN(X, Y)
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))
MIN(min(X, Y), Z) → PLUS(Y, Z)
QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) → MIN(X, Y)
The TRS R consists of the following rules:
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(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 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:
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
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,
x2)
Tags:
PLUS has argument tags [3,2,0] 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
(7) 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))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
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:
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))
MIN(s(X), s(Y)) → MIN(X, Y)
The TRS R consists of the following rules:
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
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.
MIN(min(X, Y), Z) → MIN(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:
MIN(
x0,
x1,
x2) =
MIN(
x0)
Tags:
MIN has argument tags [2,1,3] 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.
MIN(
x1,
x2) =
MIN(
x1,
x2)
min(
x1,
x2) =
min(
x1,
x2)
Z =
Z
plus(
x1,
x2) =
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:
MIN(s(X), s(Y)) → MIN(X, Y)
The TRS R consists of the following rules:
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
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.
MIN(s(X), s(Y)) → MIN(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MIN(
x0,
x1,
x2) =
MIN(
x2)
Tags:
MIN has argument tags [2,0,2] 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.
MIN(
x1,
x2) =
MIN(
x2)
s(
x1) =
s(
x1)
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:
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
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(min(X, Y), s(Y))
The TRS R consists of the following rules:
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
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(min(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 [0,0,0] 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) =
x1
s(
x1) =
s(
x1)
min(
x1,
x2) =
x1
0 =
0
Z =
Z
plus(
x1,
x2) =
plus(
x1)
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
min(X, 0) → X
min(min(X, Y), Z) → min(X, plus(Y, Z))
min(s(X), s(Y)) → min(X, Y)
(19) 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))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))
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