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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 12 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtUsableRulesProof (⇔, 0 ms)
        ↳6 CdtProblem
        ↳7 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 57 ms)
        ↳8 CdtProblem
        ↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳10 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:

h(f(x), y) → f(g(x, y))
g(x, y) → h(x, y)

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 2.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2]
transitions:
f0(0) → 0
h0(0, 0) → 1
g0(0, 0) → 2
g1(0, 0) → 3
f1(3) → 1
h1(0, 0) → 2
f1(3) → 2
h2(0, 0) → 3
f1(3) → 3

(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:

h(f(z0), z1) → f(g(z0, z1))
g(z0, z1) → h(z0, z1)
Tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))
S tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))
K tuples:none
Defined Rule Symbols:

h, g

Defined Pair Symbols:

H, G

Compound Symbols:

c, c1

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

h(f(z0), z1) → f(g(z0, z1))
g(z0, z1) → h(z0, z1)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))
S tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

H, G

Compound Symbols:

c, c1

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(G(x1, x2)) = [2] + [2]x1   
POL(H(x1, x2)) = [2]x1   
POL(c(x1)) = x1   
POL(c1(x1)) = x1   
POL(f(x1)) = [2] + x1   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))
S tuples:none
K tuples:

H(f(z0), z1) → c(G(z0, z1))
G(z0, z1) → c1(H(z0, z1))
Defined Rule Symbols:none

Defined Pair Symbols:

H, G

Compound Symbols:

c, c1

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

The set S is empty

(10) BOUNDS(1, 1)