### (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(O(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(O(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