(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → 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:
ADMIT(x, .(u, .(v, .(w, z)))) → COND(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
ADMIT(x, .(u, .(v, .(w, z)))) → ADMIT(carry(x, u, v), z)
The TRS R consists of the following rules:
admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → 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 1 SCC with 1 less node.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADMIT(x, .(u, .(v, .(w, z)))) → ADMIT(carry(x, u, v), z)
The TRS R consists of the following rules:
admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ADMIT(x, .(u, .(v, .(w, z)))) → ADMIT(carry(x, u, v), z)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
ADMIT(
x1,
x2) =
ADMIT(
x2)
Tags:
ADMIT has tags [0,0]
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:
POL(.(x1, x2)) = x1 + x2
POL(carry(x1, x2, x3)) = 1 + x2
POL(w) = 1
The following usable rules [FROCOS05] were oriented:
none
(6) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(8) TRUE