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

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

ackin(s(X), s(Y)) → u21(ackin(s(X), Y), X)
u21(ackout(X), Y) → u22(ackin(Y, X))

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

ackin(s(z0), s(z1)) → u21(ackin(s(z0), z1), z0)
u21(ackout(z0), z1) → u22(ackin(z1, z0))

Tuples:

ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))
U21(ackout(z0), z1) → c1(ACKIN(z1, z0))

S tuples:

ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))
U21(ackout(z0), z1) → c1(ACKIN(z1, z0))

K tuples:none
Defined Rule Symbols:

ackin, u21

Defined Pair Symbols:

ACKIN, U21

Compound Symbols:

c, c1

(3) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

U21(ackout(z0), z1) → c1(ACKIN(z1, z0))

We considered the (Usable) Rules:

u21(ackout(z0), z1) → u22(ackin(z1, z0))
ackin(s(z0), s(z1)) → u21(ackin(s(z0), z1), z0)

And the Tuples:

ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))
U21(ackout(z0), z1) → c1(ACKIN(z1, z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACKIN(x₁, x₂)) = [2]   
POL(U21(x₁, x₂)) = x₁   
POL(ackin(x₁, x₂)) = [2]x₁   
POL(ackout(x₁)) = [4]   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁)) = x₁   
POL(s(x₁)) = 0   
POL(u21(x₁, x₂)) = [4]x₁   
POL(u22(x₁)) = [4]   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

ackin(s(z0), s(z1)) → u21(ackin(s(z0), z1), z0)
u21(ackout(z0), z1) → u22(ackin(z1, z0))

Tuples:

ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))
U21(ackout(z0), z1) → c1(ACKIN(z1, z0))

S tuples:

ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))

K tuples:

U21(ackout(z0), z1) → c1(ACKIN(z1, z0))

Defined Rule Symbols:

ackin, u21

Defined Pair Symbols:

ACKIN, U21

Compound Symbols:

c, c1

(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))

We considered the (Usable) Rules:

u21(ackout(z0), z1) → u22(ackin(z1, z0))
ackin(s(z0), s(z1)) → u21(ackin(s(z0), z1), z0)

And the Tuples:

ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))
U21(ackout(z0), z1) → c1(ACKIN(z1, z0))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(ACKIN(x₁, x₂)) = [3] + [2]x₂   
POL(U21(x₁, x₂)) = [2]x₁   
POL(ackin(x₁, x₂)) = 0   
POL(ackout(x₁)) = [4] + x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁)) = x₁   
POL(s(x₁)) = [4] + x₁   
POL(u21(x₁, x₂)) = [4]x₁   
POL(u22(x₁)) = [5]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

ackin(s(z0), s(z1)) → u21(ackin(s(z0), z1), z0)
u21(ackout(z0), z1) → u22(ackin(z1, z0))

Tuples:

ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))
U21(ackout(z0), z1) → c1(ACKIN(z1, z0))

S tuples:none
K tuples:

U21(ackout(z0), z1) → c1(ACKIN(z1, z0))
ACKIN(s(z0), s(z1)) → c(U21(ackin(s(z0), z1), z0), ACKIN(s(z0), z1))

Defined Rule Symbols:

ackin, u21

Defined Pair Symbols:

ACKIN, U21

Compound Symbols:

c, c1

(7) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(8) BOUNDS(1, 1)