The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^1)).

(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 CpxTRS 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, .(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

(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(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(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

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(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]   

(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:none
K tuples:

ADMIT(z0, .(z1, .(z2, .(w, z3)))) → c1(ADMIT(carry(z0, z1, z2), z3))

Defined Rule Symbols:

admit, cond

Defined Pair Symbols:

ADMIT

Compound Symbols:

c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))