(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(f(x)) → f(x)
f(s(x)) → f(x)
g(s(0)) → g(f(s(0)))
Rewrite Strategy: INNERMOST
(1) NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID) transformation)
The following defined symbols can occur below the 0th argument of g: f
Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
f(f(x)) → f(x)
(2) 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(0)) → g(f(s(0)))
f(s(x)) → f(x)
Rewrite Strategy: INNERMOST
(3) 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: 3
Accept states: [4]
Transitions:
3→4[g_1|0, f_1|0, f_1|1]
3→5[g_1|1]
4→4[s_1|0, 0|0]
5→6[f_1|1]
5→7[f_1|2]
6→7[s_1|1]
7→4[0|1]
(4) BOUNDS(1, n^1)
(5) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
g(s(0)) → g(f(s(0)))
f(s(z0)) → f(z0)
Tuples:
G(s(0)) → c(G(f(s(0))), F(s(0)))
F(s(z0)) → c1(F(z0))
S tuples:
G(s(0)) → c(G(f(s(0))), F(s(0)))
F(s(z0)) → c1(F(z0))
K tuples:none
Defined Rule Symbols:
g, f
Defined Pair Symbols:
G, F
Compound Symbols:
c, c1
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
g(s(0)) → g(f(s(0)))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(s(z0)) → f(z0)
Tuples:
G(s(0)) → c(G(f(s(0))), F(s(0)))
F(s(z0)) → c1(F(z0))
S tuples:
G(s(0)) → c(G(f(s(0))), F(s(0)))
F(s(z0)) → c1(F(z0))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
G, F
Compound Symbols:
c, c1
(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
G(s(0)) → c(G(f(s(0))), F(s(0)))
F(s(z0)) → c1(F(z0))
We considered the (Usable) Rules:
f(s(z0)) → f(z0)
And the Tuples:
G(s(0)) → c(G(f(s(0))), F(s(0)))
F(s(z0)) → c1(F(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(F(x1)) = x1
POL(G(x1)) = [2]x1
POL(c(x1, x2)) = x1 + x2
POL(c1(x1)) = x1
POL(f(x1)) = 0
POL(s(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(s(z0)) → f(z0)
Tuples:
G(s(0)) → c(G(f(s(0))), F(s(0)))
F(s(z0)) → c1(F(z0))
S tuples:none
K tuples:
G(s(0)) → c(G(f(s(0))), F(s(0)))
F(s(z0)) → c1(F(z0))
Defined Rule Symbols:
f
Defined Pair Symbols:
G, F
Compound Symbols:
c, c1
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)