(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(a)) → f(g(n__f(n__a)))
f(X) → n__f(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__a) → a
activate(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 3.
The certificate found is represented by the following graph.
Start state: 7
Accept states: [8]
Transitions:
7→8[f_1|0, a|0, activate_1|0, n__f_1|1, n__a|1, a|1, g_1|1, n__a|2]
7→9[f_1|1, n__f_1|2]
7→10[f_1|2, n__f_1|3]
8→8[g_1|0, n__f_1|0, n__a|0]
9→8[activate_1|1, n__f_1|1, a|1, n__a|1, g_1|1, n__a|2]
9→9[f_1|1, n__f_1|2]
9→10[f_1|2, n__f_1|3]
10→11[g_1|2]
11→12[n__f_1|2]
12→8[n__a|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(f(a)) → f(g(n__f(n__a)))
f(z0) → n__f(z0)
a → n__a
activate(n__f(z0)) → f(activate(z0))
activate(n__a) → a
activate(z0) → z0
Tuples:
F(f(a)) → c(F(g(n__f(n__a))))
F(z0) → c1
A → c2
ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__a) → c4(A)
ACTIVATE(z0) → c5
S tuples:
F(f(a)) → c(F(g(n__f(n__a))))
F(z0) → c1
A → c2
ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__a) → c4(A)
ACTIVATE(z0) → c5
K tuples:none
Defined Rule Symbols:
f, a, activate
Defined Pair Symbols:
F, A, ACTIVATE
Compound Symbols:
c, c1, c2, c3, c4, c5
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 5 trailing nodes:
F(f(a)) → c(F(g(n__f(n__a))))
A → c2
F(z0) → c1
ACTIVATE(z0) → c5
ACTIVATE(n__a) → c4(A)
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(f(a)) → f(g(n__f(n__a)))
f(z0) → n__f(z0)
a → n__a
activate(n__f(z0)) → f(activate(z0))
activate(n__a) → a
activate(z0) → z0
Tuples:
ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c3(F(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
f, a, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c3
(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(f(a)) → f(g(n__f(n__a)))
f(z0) → n__f(z0)
a → n__a
activate(n__f(z0)) → f(activate(z0))
activate(n__a) → a
activate(z0) → z0
Tuples:
ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
f, a, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c3
(9) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(f(a)) → f(g(n__f(n__a)))
f(z0) → n__f(z0)
a → n__a
activate(n__f(z0)) → f(activate(z0))
activate(n__a) → a
activate(z0) → z0
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c3
(11) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:
ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(ACTIVATE(x1)) = x1
POL(c3(x1)) = x1
POL(n__f(x1)) = [1] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
S tuples:none
K tuples:
ACTIVATE(n__f(z0)) → c3(ACTIVATE(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c3
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)