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(g(X)) → f(X)

Rewrite Strategy: INNERMOST

(1) 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[f_1|0, f_1|1]
2→2[g_1|0]

(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(g(z0)) → f(z0)

Tuples:

F(g(z0)) → c(F(z0))

S tuples:

F(g(z0)) → c(F(z0))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(g(z0)) → f(z0)

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0)) → c(F(z0))

S tuples:

F(g(z0)) → c(F(z0))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

(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.

F(g(z0)) → c(F(z0))

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

F(g(z0)) → c(F(z0))

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

POL(F(x₁)) = x₁   
POL(c(x₁)) = x₁   
POL(g(x₁)) = [1] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0)) → c(F(z0))

S tuples:none
K tuples:

F(g(z0)) → c(F(z0))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

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

The set S is empty

(10) BOUNDS(1, 1)