(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n
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:
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))
SUM(cons(0, x), y) → SUM(x, y)
WEIGHT(cons(n, cons(m, x))) → WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
WEIGHT(cons(n, cons(m, x))) → SUM(cons(n, cons(m, x)), cons(0, x))
The TRS R consists of the following rules:
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n
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 2 SCCs with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SUM(cons(0, x), y) → SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))
The TRS R consists of the following rules:
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n
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.
SUM(cons(0, x), y) → SUM(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
SUM(
x0,
x1,
x2) =
SUM(
x0,
x1)
Tags:
SUM has argument tags [2,0,1] and root tag 0
Comparison: MS
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
SUM(
x1,
x2) =
SUM
cons(
x1,
x2) =
cons(
x2)
0 =
0
s(
x1) =
x1
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))
The TRS R consists of the following rules:
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
SUM(cons(s(n), x), cons(m, y)) → SUM(cons(n, x), cons(s(m), y))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
SUM(
x0,
x1,
x2) =
SUM(
x1)
Tags:
SUM has argument tags [3,0,0] 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.
SUM(
x1,
x2) =
SUM(
x2)
cons(
x1,
x2) =
x1
s(
x1) =
s(
x1)
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(9) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n
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) TRUE
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WEIGHT(cons(n, cons(m, x))) → WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
The TRS R consists of the following rules:
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n
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.
WEIGHT(cons(n, cons(m, x))) → WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
WEIGHT(
x0,
x1) =
WEIGHT(
x1)
Tags:
WEIGHT has argument tags [0,1] 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.
WEIGHT(
x1) =
WEIGHT
cons(
x1,
x2) =
cons(
x2)
sum(
x1,
x2) =
x2
0 =
0
s(
x1) =
s
nil =
nil
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:
sum(cons(s(n), x), cons(m, y)) → sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) → sum(x, y)
sum(nil, y) → y
weight(cons(n, cons(m, x))) → weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) → n
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