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:

and(tt, X) → 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)

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → 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)
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

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

(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.

X(z0, s(z1)) → c4(PLUS(x(z0, z1), z0), X(z0, z1))
We considered the (Usable) 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))
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)) = 0   
POL(X(x1, x2)) = x2   
POL(c2(x1)) = x1   
POL(c4(x1, x2)) = x1 + x2   
POL(plus(x1, x2)) = [2] + [2]x1   
POL(s(x1)) = [2] + x1   
POL(x(x1, x2)) = [4] + [2]x2   

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → 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)
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:

and, plus, x

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

(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(z0, s(z1)) → c2(PLUS(z0, z1))
We considered the (Usable) 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))
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)) = [1] + x2   
POL(X(x1, x2)) = [2]x2 + [2]x1·x2   
POL(c2(x1)) = x1   
POL(c4(x1, x2)) = x1 + x2   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = [1] + x1   
POL(x(x1, x2)) = 0   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(tt, z0) → 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)
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:

and, plus, x

Defined Pair Symbols:

PLUS, X

Compound Symbols:

c2, c4

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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