(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) 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:
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
The set Q consists of the following terms:
admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)
(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:
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
The set Q consists of the following terms:
admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(6) 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
The set Q consists of the following terms:
admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)
We have to consider all minimal (P,Q,R)-chains.
(7) 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.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADMIT(x, .(u, .(v, .(w, z)))) → ADMIT(carry(x, u, v), z)
R is empty.
The set Q consists of the following terms:
admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)
We have to consider all minimal (P,Q,R)-chains.
(9) 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].
admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADMIT(x, .(u, .(v, .(w, z)))) → ADMIT(carry(x, u, v), z)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) 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:
- ADMIT(x, .(u, .(v, .(w, z)))) → ADMIT(carry(x, u, v), z)
The graph contains the following edges 2 > 2
(12) TRUE