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:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, plus(z1, s(z2))) → plus(times(z0, plus(z1, times(s(z2), 0))), times(z0, s(z2)))
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))

Tuples:

TIMES(z0, plus(z1, s(z2))) → c(PLUS(times(z0, plus(z1, times(s(z2), 0))), times(z0, s(z2))), TIMES(z0, plus(z1, times(s(z2), 0))), PLUS(z1, times(s(z2), 0)), TIMES(s(z2), 0), TIMES(z0, s(z2)))
TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

S tuples:

TIMES(z0, plus(z1, s(z2))) → c(PLUS(times(z0, plus(z1, times(s(z2), 0))), times(z0, s(z2))), TIMES(z0, plus(z1, times(s(z2), 0))), PLUS(z1, times(s(z2), 0)), TIMES(s(z2), 0), TIMES(z0, s(z2)))
TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c, c2, c4

(3) CdtUnreachableProof (EQUIVALENT transformation)

The following tuples could be removed as they are not reachable from basic start terms:

TIMES(z0, plus(z1, s(z2))) → c(PLUS(times(z0, plus(z1, times(s(z2), 0))), times(z0, s(z2))), TIMES(z0, plus(z1, times(s(z2), 0))), PLUS(z1, times(s(z2), 0)), TIMES(s(z2), 0), TIMES(z0, s(z2)))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, plus(z1, s(z2))) → plus(times(z0, plus(z1, times(s(z2), 0))), times(z0, s(z2)))
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))

Tuples:

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

S tuples:

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c2, c4

(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.

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

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

POL(0) = [3]   
POL(PLUS(x₁, x₂)) = 0   
POL(TIMES(x₁, x₂)) = [2]x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = [1] + x₁   
POL(times(x₁, x₂)) = [4] + [2]x₁ + [2]x₂   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, plus(z1, s(z2))) → plus(times(z0, plus(z1, times(s(z2), 0))), times(z0, s(z2)))
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))

Tuples:

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

S tuples:

PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

K tuples:

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))

Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c2, c4

(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.

PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

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

POL(0) = 0   
POL(PLUS(x₁, x₂)) = x₂   
POL(TIMES(x₁, x₂)) = [2]x₁·x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁)) = x₁   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = [1] + x₁   
POL(times(x₁, x₂)) = 0   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(z0, plus(z1, s(z2))) → plus(times(z0, plus(z1, times(s(z2), 0))), times(z0, s(z2)))
times(z0, 0) → 0
times(z0, s(z1)) → plus(times(z0, z1), z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))

Tuples:

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

S tuples:none
K tuples:

TIMES(z0, s(z1)) → c2(PLUS(times(z0, z1), z0), TIMES(z0, z1))
PLUS(z0, s(z1)) → c4(PLUS(z0, z1))

Defined Rule Symbols:

times, plus

Defined Pair Symbols:

TIMES, PLUS

Compound Symbols:

c2, c4

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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