### (0) Obligation:

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

and(tt, X) → activate(X)
plus(N, 0) → N
plus(N, s(M)) → s(plus(N, M))
activate(X) → X

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
activate(z0) → z0
Tuples:

AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, 0) → c1
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
ACTIVATE(z0) → c3
S tuples:

AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, 0) → c1
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
ACTIVATE(z0) → c3
K tuples:none
Defined Rule Symbols:

and, plus, activate

Defined Pair Symbols:

AND, PLUS, ACTIVATE

Compound Symbols:

c, c1, c2, c3

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

Removed 3 trailing nodes:

AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, 0) → c1
ACTIVATE(z0) → c3

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
activate(z0) → z0
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:

and, plus, activate

Defined Pair Symbols:

PLUS

Compound Symbols:

c2

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
activate(z0) → z0

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

PLUS

Compound Symbols:

c2

### (7) 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.

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(PLUS(x1, x2)) = [5]x2
POL(c2(x1)) = x1
POL(s(x1)) = [1] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:none
K tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

PLUS

Compound Symbols:

c2

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

The set S is empty