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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 19 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)), 33 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:

f(S(x), x2) → f(x2, x)
f(0, x2) → 0

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]
transitions:
S0(0) → 0
00() → 0
f0(0, 0) → 1
f1(0, 0) → 1
01() → 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:

f(S(z0), z1) → f(z1, z0)
f(0, z0) → 0
Tuples:

F(S(z0), z1) → c(F(z1, z0))
F(0, z0) → c1
S tuples:

F(S(z0), z1) → c(F(z1, z0))
F(0, z0) → c1
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1

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

Removed 1 trailing nodes:

F(0, z0) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(S(z0), z1) → f(z1, z0)
f(0, z0) → 0
Tuples:

F(S(z0), z1) → c(F(z1, z0))
S tuples:

F(S(z0), z1) → c(F(z1, z0))
K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(S(z0), z1) → f(z1, z0)
f(0, z0) → 0

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(S(z0), z1) → c(F(z1, z0))
S tuples:

F(S(z0), z1) → c(F(z1, z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

(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.

F(S(z0), z1) → c(F(z1, z0))
We considered the (Usable) Rules:none
And the Tuples:

F(S(z0), z1) → c(F(z1, z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(F(x1, x2)) = x1 + x2   
POL(S(x1)) = [1] + x1   
POL(c(x1)) = x1   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(S(z0), z1) → c(F(z1, z0))
S tuples:none
K tuples:

F(S(z0), z1) → c(F(z1, z0))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

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

The set S is empty

(12) BOUNDS(1, 1)