(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(g(x)) → g(g(f(x)))
f(g(x)) → g(g(g(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 1.
The certificate found is represented by the following graph.
Start state: 1
Accept states: [2]
Transitions:
1→2[f_1|0]
1→3[g_1|1]
1→5[g_1|1]
2→2[g_1|0]
3→4[g_1|1]
4→2[f_1|1]
4→3[g_1|1]
4→5[g_1|1]
5→6[g_1|1]
6→2[g_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:
f(g(z0)) → g(g(f(z0)))
f(g(z0)) → g(g(g(z0)))
Tuples:
F(g(z0)) → c(F(z0))
F(g(z0)) → c1
S tuples:
F(g(z0)) → c(F(z0))
F(g(z0)) → c1
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c, c1
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
F(g(z0)) → c1
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(g(z0)) → g(g(f(z0)))
f(g(z0)) → g(g(g(z0)))
Tuples:
F(g(z0)) → c(F(z0))
S tuples:
F(g(z0)) → c(F(z0))
K tuples:none
Defined Rule Symbols:
f
Defined Pair Symbols:
F
Compound Symbols:
c
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(g(z0)) → g(g(f(z0)))
f(g(z0)) → g(g(g(z0)))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(g(z0)) → c(F(z0))
S tuples:
F(g(z0)) → c(F(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c
(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.
F(g(z0)) → c(F(z0))
We considered the (Usable) Rules:none
And the Tuples:
F(g(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(g(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
F(g(z0)) → c(F(z0))
S tuples:none
K tuples:
F(g(z0)) → c(F(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
F
Compound Symbols:
c
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)