The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^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(0), g(x)) → f(x, g(x))
g(s(x)) → g(x)

Rewrite Strategy: INNERMOST

(1) DependencyGraphProof (BOTH BOUNDS(ID, ID) transformation)

The following rules are not reachable from basic terms in the dependency graph and can be removed:
f(s(0), g(x)) → f(x, g(x))

(2) 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:

g(s(x)) → g(x)

Rewrite Strategy: INNERMOST

(3) CpxTrsMatchBoundsProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1.
The certificate found is represented by the following graph.

Start state: 1
Accept states: [2]
Transitions:
1→2[g_1|0, g_1|1]
2→2[s_1|0]

(4) BOUNDS(1, n^1)

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

Converted Cpx (relative) TRS to CDT

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

g(s(z0)) → g(z0)

Tuples:

G(s(z0)) → c(G(z0))

S tuples:

G(s(z0)) → c(G(z0))

K tuples:none
Defined Rule Symbols:

g

Defined Pair Symbols:

G

Compound Symbols:

c

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

g(s(z0)) → g(z0)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(s(z0)) → c(G(z0))

S tuples:

G(s(z0)) → c(G(z0))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G

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.

G(s(z0)) → c(G(z0))

We considered the (Usable) Rules:none
And the Tuples:

G(s(z0)) → c(G(z0))

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

POL(G(x₁)) = x₁   
POL(c(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(s(z0)) → c(G(z0))

S tuples:none
K tuples:

G(s(z0)) → c(G(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

G

Compound Symbols:

c

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

The set S is empty

(12) BOUNDS(1, 1)