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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 14 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 17 ms)
        ↳10 CdtProblem
        ↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳12 BOUNDS(1, 1)

(0) Obligation:

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


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))
activate(X) → X

Rewrite Strategy: INNERMOST

(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 1.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3]
transitions:
tt0() → 0
00() → 0
s0(0) → 0
and0(0, 0) → 1
plus0(0, 0) → 2
activate0(0) → 3
activate1(0) → 1
plus1(0, 0) → 4
s1(4) → 2
s1(4) → 4
0 → 2
0 → 3
0 → 4
0 → 1

(2) BOUNDS(1, n^1)

(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))
activate(z0) → z0
Tuples:

AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, 0) → c1
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
ACTIVATE(z0) → c3
S tuples:

AND(tt, z0) → c(ACTIVATE(z0))
PLUS(z0, 0) → c1
PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
ACTIVATE(z0) → c3
K tuples:none
Defined Rule Symbols:

and, plus, activate

Defined Pair Symbols:

AND, PLUS, ACTIVATE

Compound Symbols:

c, c1, c2, c3

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

Removed 3 trailing nodes:

ACTIVATE(z0) → c3
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))
activate(z0) → z0
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:

and, plus, activate

Defined Pair Symbols:

PLUS

Compound Symbols:

c2

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

and(tt, z0) → activate(z0)
plus(z0, 0) → z0
plus(z0, s(z1)) → s(plus(z0, z1))
activate(z0) → z0

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

PLUS

Compound Symbols:

c2

(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.

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

POL(PLUS(x1, x2)) = x2   
POL(c2(x1)) = x1   
POL(s(x1)) = [1] + x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
S tuples:none
K tuples:

PLUS(z0, s(z1)) → c2(PLUS(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

PLUS

Compound Symbols:

c2

(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(12) BOUNDS(1, 1)