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(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), 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 1.

The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2]
transitions:
c0(0, 0) → 0
s0(0) → 0
f0(0) → 1
g0(0) → 2
s1(0) → 4
c1(0, 4) → 3
f1(3) → 1
s1(0) → 6
c1(6, 0) → 5
g1(5) → 2
s1(4) → 4
s1(6) → 6

(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(c(s(z0), z1)) → f(c(z0, s(z1)))
g(c(z0, s(z1))) → g(c(s(z0), z1))

Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(c(s(z0), z1)) → f(c(z0, s(z1)))
g(c(z0, s(z1))) → g(c(s(z0), z1))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(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.

G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

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

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

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

POL(F(x₁)) = 0   
POL(G(x₁)) = [2]x₁   
POL(c(x₁, x₂)) = x₂   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [2] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

S tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))

K tuples:

G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(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(c(s(z0), z1)) → c1(F(c(z0, s(z1))))

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

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

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

POL(F(x₁)) = x₁   
POL(G(x₁)) = 0   
POL(c(x₁, x₂)) = x₁   
POL(c1(x₁)) = x₁   
POL(c2(x₁)) = x₁   
POL(s(x₁)) = [1] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))
G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))

S tuples:none
K tuples:

G(c(z0, s(z1))) → c2(G(c(s(z0), z1)))
F(c(s(z0), z1)) → c1(F(c(z0, s(z1))))

Defined Rule Symbols:none

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

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

The set S is empty

(12) BOUNDS(1, 1)