(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(s(x)) → s(s(f(p(s(x)))))
f(0) → 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, s_1|1]
1→3[s_1|1]
2→2[s_1|0, 0|0]
3→4[s_1|1]
4→5[f_1|1]
4→7[s_1|2]
4→2[0|2]
5→6[p_1|1]
5→2[s_1|1, 0|1]
6→2[s_1|1]
7→8[s_1|2]
8→9[f_1|2]
8→7[s_1|2]
8→2[0|2]
9→10[p_1|2]
9→2[s_1|1, 0|1]
10→2[s_1|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(s(z0)) → s(s(f(p(s(z0)))))
f(0) → 0
p(s(z0)) → z0
Tuples:
F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
F(0) → c1
P(s(z0)) → c2
S tuples:
F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
F(0) → c1
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:
P(s(z0)) → c2
F(0) → c1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(s(z0)) → s(s(f(p(s(z0)))))
f(0) → 0
p(s(z0)) → z0
Tuples:
F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
S tuples:
F(s(z0)) → c(F(p(s(z0))), P(s(z0)))
K tuples:none
Defined Rule Symbols:
f, p
Defined Pair Symbols:
F
Compound Symbols:
c
(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(s(z0)) → s(s(f(p(s(z0)))))
f(0) → 0
p(s(z0)) → z0
Tuples:
F(s(z0)) → c(F(p(s(z0))))
S tuples:
F(s(z0)) → c(F(p(s(z0))))
K tuples:none
Defined Rule Symbols:
f, p
Defined Pair Symbols:
F
Compound Symbols:
c
(9) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(s(z0)) → s(s(f(p(s(z0)))))
f(0) → 0
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(s(z0)) → z0
Tuples:
F(s(z0)) → c(F(p(s(z0))))
S tuples:
F(s(z0)) → c(F(p(s(z0))))
K tuples:none
Defined Rule Symbols:
p
Defined Pair Symbols:
F
Compound Symbols:
c
(11) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace
F(
s(
z0)) →
c(
F(
p(
s(
z0)))) by
F(s(z0)) → c(F(z0))
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(s(z0)) → z0
Tuples:
F(s(z0)) → c(F(z0))
S tuples:
F(s(z0)) → c(F(z0))
K tuples:none
Defined Rule Symbols:
p
Defined Pair Symbols:
F
Compound Symbols:
c
(13) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
p(s(z0)) → z0
(14) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(s(z0)) → c(F(z0))
S tuples:
F(s(z0)) → c(F(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c
(15) 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(z0)) → c(F(z0))
We considered the (Usable) Rules:none
And the Tuples:
F(s(z0)) → c(F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1)) = x1
POL(c(x1)) = x1
POL(s(x1)) = [1] + x1
(16) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(s(z0)) → c(F(z0))
S tuples:none
K tuples:
F(s(z0)) → c(F(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c
(17) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(18) BOUNDS(1, 1)