The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^1).

0 CpxTRS
    ↳1 CpxTrsMatchBoundsTAProof (⇔, 21 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (ComplexityIfPolyImplication, 0 ms)
        ↳6 CdtProblem
        ↳7 CdtUsableRulesProof (⇔, 0 ms)
        ↳8 CdtProblem
        ↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 65 ms)
        ↳10 CdtProblem
        ↳11 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳12 BOUNDS(1, 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(X, s(Y))) → f(c(s(X), Y))
g(c(s(X), Y)) → f(c(X, s(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:
c0(0, 0) → 0
s0(0) → 0
f0(0) → 1
g0(0) → 2
s1(0) → 4
c1(4, 0) → 3
f1(3) → 1
s1(0) → 6
c1(0, 6) → 5
f1(5) → 2
s1(4) → 4
s2(0) → 8
c2(8, 0) → 7
f2(7) → 2
s1(8) → 4
f1(3) → 2

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

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

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
G(c(s(z0), z1)) → c2(F(c(z0, s(z1))))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F, G

Compound Symbols:

c1, c2

(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)

Removed 1 leading nodes:

G(c(s(z0), z1)) → c2(F(c(z0, s(z1))))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
S tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
K tuples:none
Defined Rule Symbols:

f, g

Defined Pair Symbols:

F

Compound Symbols:

c1

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
S tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c1

(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(z0, s(z1))) → c1(F(c(s(z0), z1)))
We considered the (Usable) Rules:none
And the Tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :

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

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
S tuples:none
K tuples:

F(c(z0, s(z1))) → c1(F(c(s(z0), z1)))
Defined Rule Symbols:none

Defined Pair Symbols:

F

Compound Symbols:

c1

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

The set S is empty

(12) BOUNDS(1, 1)