### (0) Obligation:

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

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: FULL

### (1) RcToIrcProof (BOTH BOUNDS(ID, ID) transformation)

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
x([], s(M))

The defined contexts are:
plus([], x1)

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

### (2) Obligation:

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

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

### (3) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

### (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:

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

### (5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 4 trailing nodes:

X(z0, 0) → c3
ACTIVATE(z0) → c5
PLUS(z0, 0) → c1
AND(tt, z0) → c(ACTIVATE(z0))

### (6) 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

### (7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

and(tt, z0) → activate(z0)
activate(z0) → z0

### (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))
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

### (9) CdtRuleRemovalProof (UPPER BOUND(ADD(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) = 0
POL(PLUS(x1, x2)) = 0
POL(X(x1, x2)) = 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

### (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:

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

### (11) CdtRuleRemovalProof (UPPER BOUND(ADD(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) = [1]
POL(PLUS(x1, x2)) = [2] + x2
POL(X(x1, x2)) = x2 + x1·x2
POL(c2(x1)) = x1
POL(c4(x1, x2)) = x1 + x2
POL(plus(x1, x2)) = [2]x1 + x2 + x22 + [2]x1·x2 + x12
POL(s(x1)) = [2] + x1
POL(x(x1, x2)) = [1] + x1 + x2 + x22 + [2]x1·x2

### (12) 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

### (13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty