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:

U11(tt, M, N) → U12(tt, activate(M), activate(N))
U12(tt, M, N) → s(plus(activate(N), activate(M)))
U21(tt, M, N) → U22(tt, activate(M), activate(N))
U22(tt, M, N) → plus(x(activate(N), activate(M)), activate(N))
plus(N, 0) → N
plus(N, s(M)) → U11(tt, M, N)
x(N, 0) → 0
x(N, s(M)) → U21(tt, M, N)
activate(X) → X

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0, z1) → U12(tt, activate(z0), activate(z1))
U12(tt, z0, z1) → s(plus(activate(z1), activate(z0)))
U21(tt, z0, z1) → U22(tt, activate(z0), activate(z1))
U22(tt, z0, z1) → plus(x(activate(z1), activate(z0)), activate(z1))
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
activate(z0) → z0

Tuples:

U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)), ACTIVATE(z1), ACTIVATE(z0))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)), ACTIVATE(z1), ACTIVATE(z0), ACTIVATE(z1))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))

S tuples:

U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)), ACTIVATE(z1), ACTIVATE(z0))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)), ACTIVATE(z0), ACTIVATE(z1))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)), ACTIVATE(z1), ACTIVATE(z0), ACTIVATE(z1))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))

K tuples:none
Defined Rule Symbols:

U11, U12, U21, U22, plus, x, activate

Defined Pair Symbols:

U11', U12', U21', U22', PLUS, X

Compound Symbols:

c, c1, c2, c3, c5, c7

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

Removed 9 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0, z1) → U12(tt, activate(z0), activate(z1))
U12(tt, z0, z1) → s(plus(activate(z1), activate(z0)))
U21(tt, z0, z1) → U22(tt, activate(z0), activate(z1))
U22(tt, z0, z1) → plus(x(activate(z1), activate(z0)), activate(z1))
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
activate(z0) → z0

Tuples:

PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))

S tuples:

PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))

K tuples:none
Defined Rule Symbols:

U11, U12, U21, U22, plus, x, activate

Defined Pair Symbols:

PLUS, X, U11', U12', U21', U22'

Compound Symbols:

c5, c7, c, c1, c2, c3

(5) CdtPolyRedPairProof (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)) → c7(U21'(tt, z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))

We considered the (Usable) Rules:

activate(z0) → z0
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
U21(tt, z0, z1) → U22(tt, activate(z0), activate(z1))
U22(tt, z0, z1) → plus(x(activate(z1), activate(z0)), activate(z1))
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
U11(tt, z0, z1) → U12(tt, activate(z0), activate(z1))
U12(tt, z0, z1) → s(plus(activate(z1), activate(z0)))

And the Tuples:

PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))

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

POL(0) = [4]   
POL(PLUS(x₁, x₂)) = 0   
POL(U11(x₁, x₂, x₃)) = [3] + [3]x₁ + [3]x₂ + [3]x₃   
POL(U11'(x₁, x₂, x₃)) = 0   
POL(U12(x₁, x₂, x₃)) = [3] + [3]x₁ + [3]x₂ + [3]x₃   
POL(U12'(x₁, x₂, x₃)) = x₁   
POL(U21(x₁, x₂, x₃)) = [5] + x₁ + [2]x₂   
POL(U21'(x₁, x₂, x₃)) = [2] + [4]x₂   
POL(U22(x₁, x₂, x₃)) = [3] + [2]x₁ + [5]x₂ + [3]x₃   
POL(U22'(x₁, x₂, x₃)) = [1] + [4]x₂   
POL(X(x₁, x₂)) = [4]x₂   
POL(activate(x₁)) = x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(plus(x₁, x₂)) = [3] + [3]x₁ + [5]x₂   
POL(s(x₁)) = [1] + x₁   
POL(tt) = 0   
POL(x(x₁, x₂)) = [4]   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0, z1) → U12(tt, activate(z0), activate(z1))
U12(tt, z0, z1) → s(plus(activate(z1), activate(z0)))
U21(tt, z0, z1) → U22(tt, activate(z0), activate(z1))
U22(tt, z0, z1) → plus(x(activate(z1), activate(z0)), activate(z1))
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
activate(z0) → z0

Tuples:

PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))

S tuples:

PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))

K tuples:

X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))

Defined Rule Symbols:

U11, U12, U21, U22, plus, x, activate

Defined Pair Symbols:

PLUS, X, U11', U12', U21', U22'

Compound Symbols:

c5, c7, c, c1, c2, c3

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

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

U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))

We considered the (Usable) Rules:

activate(z0) → z0
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
U21(tt, z0, z1) → U22(tt, activate(z0), activate(z1))
U22(tt, z0, z1) → plus(x(activate(z1), activate(z0)), activate(z1))
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
U11(tt, z0, z1) → U12(tt, activate(z0), activate(z1))
U12(tt, z0, z1) → s(plus(activate(z1), activate(z0)))

And the Tuples:

PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))

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

POL(0) = 0   
POL(PLUS(x₁, x₂)) = [2]x₂   
POL(U11(x₁, x₂, x₃)) = 0   
POL(U11'(x₁, x₂, x₃)) = [2]x₁ + [2]x₂   
POL(U12(x₁, x₂, x₃)) = 0   
POL(U12'(x₁, x₂, x₃)) = [2]x₂   
POL(U21(x₁, x₂, x₃)) = 0   
POL(U21'(x₁, x₂, x₃)) = [2] + [3]x₁ + x₂ + [2]x₃ + [3]x₂·x₃ + [3]x₁·x₃ + [2]x₁·x₂ + [2]x₂²   
POL(U22(x₁, x₂, x₃)) = 0   
POL(U22'(x₁, x₂, x₃)) = x₁ + x₂ + [2]x₃ + [3]x₂·x₃ + [3]x₁·x₃ + [2]x₁·x₂ + [2]x₂²   
POL(X(x₁, x₂)) = [3]x₁ + [2]x₂² + [3]x₁·x₂   
POL(activate(x₁)) = x₁   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(c3(x₁, x₂)) = x₁ + x₂   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = [2] + x₁   
POL(tt) = [2]   
POL(x(x₁, x₂)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

U11(tt, z0, z1) → U12(tt, activate(z0), activate(z1))
U12(tt, z0, z1) → s(plus(activate(z1), activate(z0)))
U21(tt, z0, z1) → U22(tt, activate(z0), activate(z1))
U22(tt, z0, z1) → plus(x(activate(z1), activate(z0)), activate(z1))
plus(z0, 0) → z0
plus(z0, s(z1)) → U11(tt, z1, z0)
x(z0, 0) → 0
x(z0, s(z1)) → U21(tt, z1, z0)
activate(z0) → z0

Tuples:

PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))

S tuples:

PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))

K tuples:

X(z0, s(z1)) → c7(U21'(tt, z1, z0))
U21'(tt, z0, z1) → c2(U22'(tt, activate(z0), activate(z1)))
U22'(tt, z0, z1) → c3(PLUS(x(activate(z1), activate(z0)), activate(z1)), X(activate(z1), activate(z0)))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))

Defined Rule Symbols:

U11, U12, U21, U22, plus, x, activate

Defined Pair Symbols:

PLUS, X, U11', U12', U21', U22'

Compound Symbols:

c5, c7, c, c1, c2, c3

(9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

U12'(tt, z0, z1) → c1(PLUS(activate(z1), activate(z0)))
PLUS(z0, s(z1)) → c5(U11'(tt, z1, z0))
U11'(tt, z0, z1) → c(U12'(tt, activate(z0), activate(z1)))

Now S is empty

(10) BOUNDS(O(1), O(1))