(0) Obligation:

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


The TRS R consists of the following rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(X)) → 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, p_1|0, 0|1, cons_1|1, s_1|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(z0)) → z0
Tuples:

F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
P(s(z0)) → c2
S tuples:

F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
P(s(z0)) → 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:

F(0) → c
P(s(z0)) → c2

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

f(0) → cons(0)
f(s(0)) → f(p(s(0)))
p(s(z0)) → z0
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(z0)) → z0
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(z0)) → z0
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) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)

Use narrowing to replace F(s(0)) → c1(F(p(s(0)))) by

F(s(0)) → c1(F(0))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
Tuples:

F(s(0)) → c1(F(0))
S tuples:

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

p

Defined Pair Symbols:

F

Compound Symbols:

c1

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

Removed 1 trailing nodes:

F(s(0)) → c1(F(0))

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

p(s(z0)) → z0
Tuples:none
S tuples:none
K tuples:none
Defined Rule Symbols:

p

Defined Pair Symbols:none

Compound Symbols:none

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

The set S is empty

(16) BOUNDS(1, 1)