(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