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

0 CpxTRS
    ↳1 CpxTrsMatchBoundsProof (⇔, 9 ms)
    ↳2 BOUNDS(1, n^1)
    ↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
    ↳4 CdtProblem
        ↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳6 CdtProblem
        ↳7 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳8 CdtProblem
        ↳9 CdtUsableRulesProof (⇔, 0 ms)
        ↳10 CdtProblem
        ↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 65 ms)
        ↳12 CdtProblem
        ↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
        ↳14 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(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

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, p_1|0, 0|1]
1→3[cons_1|1]
1→4[f_1|1]
1→7[cons_1|2]
2→2[0|0, cons_1|0, s_1|0]
3→2[0|1]
4→5[p_1|1]
4→2[0|2]
5→6[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) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Tuples:

F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
P(s(0)) → c2
S tuples:

F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
P(s(0)) → c2
K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F, P

Compound Symbols:

c, c1, c2

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

Removed 2 trailing nodes:

P(s(0)) → c2
F(0) → c

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))
S tuples:

F(s(0)) → c1(F(p(s(0))), P(s(0)))
K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c1

(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing tuple parts

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:

F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:

f, p

Defined Pair Symbols:

F

Compound Symbols:

c1

(9) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(0)) → 0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:

F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:

F

Compound Symbols:

c1

(11) 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(s(0)) → c1(F(p(s(0))))
We considered the (Usable) Rules:

p(s(0)) → 0
And the Tuples:

F(s(0)) → c1(F(p(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]   
POL(F(x1)) = x1   
POL(c1(x1)) = x1   
POL(p(x1)) = [1]   
POL(s(x1)) = [1] + x1   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(0)) → 0
Tuples:

F(s(0)) → c1(F(p(s(0))))
S tuples:none
K tuples:

F(s(0)) → c1(F(p(s(0))))
Defined Rule Symbols:

p

Defined Pair Symbols:

F

Compound Symbols:

c1

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

The set S is empty

(14) BOUNDS(1, 1)