(0) Obligation:
Runtime Complexity TRS:
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
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
admit(z0, nil) → nil
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3)))))
cond(true, z0) → z0
Tuples:
ADMIT(z0, nil) → c
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))
COND(true, z0) → c2
S tuples:
ADMIT(z0, nil) → c
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))
COND(true, z0) → c2
K tuples:none
Defined Rule Symbols:
admit, cond
Defined Pair Symbols:
ADMIT, COND
Compound Symbols:
c, c1, c2
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
ADMIT(z0, nil) → c
COND(true, z0) → c2
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
admit(z0, nil) → nil
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3)))))
cond(true, z0) → z0
Tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))
S tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(COND(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3))))), ADMIT(carry(z0, z1, z2), z3))
K tuples:none
Defined Rule Symbols:
admit, cond
Defined Pair Symbols:
ADMIT
Compound Symbols:
c1
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
admit(z0, nil) → nil
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3)))))
cond(true, z0) → z0
Tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
S tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
K tuples:none
Defined Rule Symbols:
admit, cond
Defined Pair Symbols:
ADMIT
Compound Symbols:
c1
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
admit(z0, nil) → nil
admit(z0, .(z1, .(z2, .(w, z3)))) → cond(=(sum(z0, z1, z2), w), .(z1, .(z2, .(w, admit(carry(z0, z1, z2), z3)))))
cond(true, z0) → z0
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
S tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
ADMIT
Compound Symbols:
c1
(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
We considered the (Usable) Rules:none
And the Tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(.(x1, x2)) = [4] + x1 + x2
POL(ADMIT(x1, x2)) = [2]x1 + [2]x2
POL(c1(x1)) = x1
POL(carry(x1, x2, x3)) = [3]
POL(w) = [4]
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
S tuples:none
K tuples:
ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))
Defined Rule Symbols:none
Defined Pair Symbols:
ADMIT
Compound Symbols:
c1
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)