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), y, y) → g(f(x, x, y))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(g(z0), z1, z1) → g(f(z0, z0, z1))

Tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))

S tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c

(3) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(g(z0), z1, z1) → g(f(z0, z0, z1))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))

S tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

(5) 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), z1, z1) → c(F(z0, z0, z1))

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

F(g(z0), z1, z1) → c(F(z0, z0, z1))

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

POL(F(x₁, x₂, x₃)) = x₁   
POL(c(x₁)) = x₁   
POL(g(x₁)) = [1] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))

S tuples:none
K tuples:

F(g(z0), z1, z1) → c(F(z0, z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c

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

The set S is empty

(8) BOUNDS(1, 1)