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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
↳2 CdtProblem
↳3 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))))
↳4 CdtProblem
↳5 CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))))
↳6 CdtProblem
↳7 SIsEmptyProof (⇔)
↳8 BOUNDS(O(1), O(1))

(0) Obligation:

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

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:

PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:

PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
K tuples:none
Defined Rule Symbols:

plus, times

Defined Pair Symbols:

PLUS, TIMES

Compound Symbols:

c, c1

(3) 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.

TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
We considered the (Usable) Rules:

times(z0, s(z1)) → plus(z0, times(z1, z0))
plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
And the Tuples:

PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(PLUS(x1, x2)) = 0   
POL(TIMES(x1, x2)) = x1 + x2   
POL(c(x1, x2)) = x1 + x2   
POL(c1(x1, x2)) = x1 + x2   
POL(plus(x1, x2)) = [3]   
POL(s(x1)) = [1] + x1   
POL(times(x1, x2)) = [3]   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:

PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:

PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
K tuples:

TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
Defined Rule Symbols:

plus, times

Defined Pair Symbols:

PLUS, TIMES

Compound Symbols:

c, c1

(5) 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.

PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
We considered the (Usable) Rules:

times(z0, s(z1)) → plus(z0, times(z1, z0))
plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
And the Tuples:

PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(PLUS(x1, x2)) = x1   
POL(TIMES(x1, x2)) = [2]x1·x2   
POL(c(x1, x2)) = x1 + x2   
POL(c1(x1, x2)) = x1 + x2   
POL(plus(x1, x2)) = [2] + [2]x1 + x2   
POL(s(x1)) = [3] + x1   
POL(times(x1, x2)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(plus(z0, z1), z2) → plus(z0, plus(z1, z2))
times(z0, s(z1)) → plus(z0, times(z1, z0))
Tuples:

PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
S tuples:none
K tuples:

TIMES(z0, s(z1)) → c1(PLUS(z0, times(z1, z0)), TIMES(z1, z0))
PLUS(plus(z0, z1), z2) → c(PLUS(z0, plus(z1, z2)), PLUS(z1, z2))
Defined Rule Symbols:

plus, times

Defined Pair Symbols:

PLUS, TIMES

Compound Symbols:

c, c1

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))