The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(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))
x(N, 0) → 0
x(N, s(M)) → plus(x(N, M), N)
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))
x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
activate(z0) → z0

Tuples:

AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, 0) → c1
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, 0) → c3
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
ACTIVATE(z0) → c5

S tuples:

AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, 0) → c1
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, 0) → c3
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
ACTIVATE(z0) → c5

K tuples:none
Defined Rule Symbols:

and, plus, x, activate

Defined Pair Symbols:

AND, PLUS, X, ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5

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

Removed 4 trailing nodes:

AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, 0) → c1
ACTIVATE(z0) → c5
X(z0, 0) → 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))
x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
activate(z0) → z0

Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

K tuples:none
Defined Rule Symbols:

and, plus, x, activate

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

and(tt, z0) → activate(z0)
activate(z0) → z0

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))

Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

K tuples:none
Defined Rule Symbols:

x, plus

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

(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.

X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

We considered the (Usable) Rules:none
And the Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

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

POL(0) = [3]   
POL(PLUS(x₁, x₂)) = [5]   
POL(X(x₁, x₂)) = [4]x₂   
POL(c2(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(plus(x₁, x₂)) = [2] + [3]x₁ + [2]x₂   
POL(s(x₁)) = [4] + x₁   
POL(x(x₁, x₂)) = [4] + [2]x₁ + [3]x₂   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))

Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))

K tuples:

X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

Defined Rule Symbols:

x, plus

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

(9) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) 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))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

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

POL(0) = 0   
POL(PLUS(x₁, x₂)) = [3] + [2]x₂   
POL(X(x₁, x₂)) = [2]x₂ + x₁·x₂   
POL(c2(x₁)) = x₁   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = [2] + x₁   
POL(x(x₁, x₂)) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

x(z0, 0) → 0
x(z0, s(z1)) → plus(x(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))

Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))

S tuples:none
K tuples:

X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))

Defined Rule Symbols:

x, plus

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

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

The set S is empty

(12) BOUNDS(O(1), O(1))