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(f(0, x), 1) → f(g(f(x, x)), x)
f(g(x), y) → g(f(x, y))

Rewrite Strategy: INNERMOST

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

The TRS does not nest defined symbols.
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
f(f(0, x), 1) → f(g(f(x, x)), 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:

f(g(x), y) → g(f(x, y))

Rewrite Strategy: INNERMOST

(3) 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:
g0(0) → 0
f0(0, 0) → 1
f1(0, 0) → 2
g1(2) → 1
g1(2) → 2

(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:

f(g(z0), z1) → g(f(z0, z1))

Tuples:

F(g(z0), z1) → c(F(z0, z1))

S tuples:

F(g(z0), z1) → c(F(z0, z1))

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(g(z0), z1) → g(f(z0, z1))

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0), z1) → c(F(z0, z1))

S tuples:

F(g(z0), z1) → c(F(z0, z1))

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(g(z0), z1) → c(F(z0, z1))

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

F(g(z0), z1) → c(F(z0, z1))

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

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

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(g(z0), z1) → c(F(z0, z1))

S tuples:none
K tuples:

F(g(z0), z1) → c(F(z0, z1))

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)