(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:
g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(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 1.
The certificate found is represented by the following graph.
Start state: 1
Accept states: [2]
Transitions:
1→2[g_1|0, f_1|0, f_1|1, 0|1]
1→3[s_1|1]
1→4[s_1|1]
2→2[s_1|0, 0|0]
3→2[0|1]
4→5[s_1|1]
5→2[g_1|1, f_1|1, 0|1]
5→3[s_1|1]
5→4[s_1|1]
(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:
g(s(z0)) → f(z0)
g(0) → 0
f(0) → s(0)
f(s(z0)) → s(s(g(z0)))
Tuples:
G(s(z0)) → c(F(z0))
G(0) → c1
F(0) → c2
F(s(z0)) → c3(G(z0))
S tuples:
G(s(z0)) → c(F(z0))
G(0) → c1
F(0) → c2
F(s(z0)) → c3(G(z0))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c, c1, c2, c3
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing nodes:
G(0) → c1
F(0) → c2
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(s(z0)) → f(z0)
g(0) → 0
f(0) → s(0)
f(s(z0)) → s(s(g(z0)))
Tuples:
G(s(z0)) → c(F(z0))
F(s(z0)) → c3(G(z0))
S tuples:
G(s(z0)) → c(F(z0))
F(s(z0)) → c3(G(z0))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c, c3
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
g(s(z0)) → f(z0)
g(0) → 0
f(0) → s(0)
f(s(z0)) → s(s(g(z0)))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
G(s(z0)) → c(F(z0))
F(s(z0)) → c3(G(z0))
S tuples:
G(s(z0)) → c(F(z0))
F(s(z0)) → c3(G(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
G, F
Compound Symbols:
c, c3
(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
G(s(z0)) → c(F(z0))
F(s(z0)) → c3(G(z0))
We considered the (Usable) Rules:none
And the Tuples:
G(s(z0)) → c(F(z0))
F(s(z0)) → c3(G(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(F(x1)) = [2]x1 + [2]x12
POL(G(x1)) = [2] + [2]x12
POL(c(x1)) = x1
POL(c3(x1)) = x1
POL(s(x1)) = [2] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
G(s(z0)) → c(F(z0))
F(s(z0)) → c3(G(z0))
S tuples:none
K tuples:
G(s(z0)) → c(F(z0))
F(s(z0)) → c3(G(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
G, F
Compound Symbols:
c, c3
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)