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(0) → s(0)
f(s(0)) → s(0)
f(s(s(x))) → f(f(s(x)))

Rewrite Strategy: INNERMOST

(1) CpxTrsMatchBoundsProof (EQUIVALENT transformation)

A linear upper bound on the runtime complexity of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 2.
The certificate found is represented by the following graph.

Start state: 1
Accept states: [2]
Transitions:
1→2[f_1|0]
1→3[s_1|1]
1→4[f_1|1]
1→7[s_1|2]
2→2[0|0, s_1|0]
3→2[0|1]
4→5[f_1|1]
4→6[s_1|1]
4→4[f_1|1]
4→7[s_1|2]
5→2[s_1|1]
6→2[0|1]
7→2[0|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(0) → s(0)
f(s(0)) → s(0)
f(s(s(z0))) → f(f(s(z0)))

Tuples:

F(0) → c
F(s(0)) → c1
F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

S tuples:

F(0) → c
F(s(0)) → c1
F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c, c1, c2

(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 2 trailing nodes:

F(s(0)) → c1
F(0) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → s(0)
f(s(0)) → s(0)
f(s(s(z0))) → f(f(s(z0)))

Tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

S tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

K tuples:none
Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2

(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.

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

We considered the (Usable) Rules:

f(0) → s(0)
f(s(s(z0))) → f(f(s(z0)))
f(s(0)) → s(0)

And the Tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

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

POL(0) = 0   
POL(F(x₁)) = [1] + [2]x₁ + x₁²   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(f(x₁)) = [2]   
POL(s(x₁)) = [2] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → s(0)
f(s(0)) → s(0)
f(s(s(z0))) → f(f(s(z0)))

Tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

S tuples:none
K tuples:

F(s(s(z0))) → c2(F(f(s(z0))), F(s(z0)))

Defined Rule Symbols:

f

Defined Pair Symbols:

F

Compound Symbols:

c2

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

The set S is empty

(10) BOUNDS(1, 1)