(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:
a(b(x)) → b(b(a(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[a_1|0]
1→3[b_1|1]
2→2[b_1|0]
3→4[b_1|1]
4→2[a_1|1]
4→3[b_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:
a(b(z0)) → b(b(a(z0)))
Tuples:
A(b(z0)) → c(A(z0))
S tuples:
A(b(z0)) → c(A(z0))
K tuples:none
Defined Rule Symbols:
a
Defined Pair Symbols:
A
Compound Symbols:
c
(5) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
a(b(z0)) → b(b(a(z0)))
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
A(b(z0)) → c(A(z0))
S tuples:
A(b(z0)) → c(A(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
A
Compound Symbols:
c
(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
A(b(z0)) → c(A(z0))
We considered the (Usable) Rules:none
And the Tuples:
A(b(z0)) → c(A(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(A(x1)) = x1
POL(b(x1)) = [1] + x1
POL(c(x1)) = x1
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
A(b(z0)) → c(A(z0))
S tuples:none
K tuples:
A(b(z0)) → c(A(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
A
Compound Symbols:
c
(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(10) BOUNDS(1, 1)