### (0) Obligation:

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

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: FULL

### (1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
ackin(s([]), s(Y))

The defined contexts are:
u21([], x1)
ackin(x0, [])
ackin(s(x0), [])

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

### (2) Obligation:

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

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

### (3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

### (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))
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

### (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.

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)) = 0
POL(U21(x1, x2)) = x1
POL(ackin(x1, x2)) = 0
POL(ackout(x1)) = [1]
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(s(x1)) = 0
POL(u21(x1, x2)) = 0
POL(u22(x1)) = 0

### (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:

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

### (7) 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)) = x2
POL(U21(x1, x2)) = x1
POL(ackin(x1, x2)) = 0
POL(ackout(x1)) = [1] + x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(s(x1)) = [1] + x1
POL(u21(x1, x2)) = x1
POL(u22(x1)) = 0

### (8) 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

### (9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty