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:

g(f(x), y) → f(h(x, y))
h(x, y) → g(x, f(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
g0(0, 0) → 1
h0(0, 0) → 2
h1(0, 0) → 3
f1(3) → 1
f1(0) → 4
g1(0, 4) → 2
h1(0, 4) → 3
f1(3) → 2
f2(0) → 5
g2(0, 5) → 3
f2(4) → 5
h1(0, 5) → 3
f1(3) → 3
f2(5) → 5

(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:

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

Tuples:

G(f(z0), z1) → c(H(z0, z1))
H(z0, z1) → c1(G(z0, f(z1)))

S tuples:

G(f(z0), z1) → c(H(z0, z1))
H(z0, z1) → c1(G(z0, f(z1)))

K tuples:none
Defined Rule Symbols:

g, h

Defined Pair Symbols:

G, H

Compound Symbols:

c, c1

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(f(z0), z1) → c(H(z0, z1))
H(z0, z1) → c1(G(z0, f(z1)))

S tuples:

G(f(z0), z1) → c(H(z0, z1))
H(z0, z1) → c1(G(z0, f(z1)))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

G, H

Compound Symbols:

c, c1

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

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

G(f(z0), z1) → c(H(z0, z1))

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

G(f(z0), z1) → c(H(z0, z1))
H(z0, z1) → c1(G(z0, f(z1)))

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

POL(G(x₁, x₂)) = x₁ + x₁²   
POL(H(x₁, x₂)) = [2]x₁ + x₁²   
POL(c(x₁)) = x₁   
POL(c1(x₁)) = x₁   
POL(f(x₁)) = [2] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

G(f(z0), z1) → c(H(z0, z1))
H(z0, z1) → c1(G(z0, f(z1)))

S tuples:

H(z0, z1) → c1(G(z0, f(z1)))

K tuples:

G(f(z0), z1) → c(H(z0, z1))

Defined Rule Symbols:none

Defined Pair Symbols:

G, H

Compound Symbols:

c, c1

(9) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

H(z0, z1) → c1(G(z0, f(z1)))
G(f(z0), z1) → c(H(z0, z1))

Now S is empty

(10) BOUNDS(1, 1)