(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(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X
Rewrite Strategy: INNERMOST
(1) CpxTrsMatchBoundsTAProof (EQUIVALENT transformation)
A linear upper bound on the runtime complexity of the TRS R could be shown with a Match-Bound[TAB_LEFTLINEAR,TAB_NONLEFTLINEAR] (for contructor-based start-terms) of 2.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3]
transitions:
00() → 0
cons0(0, 0) → 0
n__f0(0) → 0
s0(0) → 0
f0(0) → 1
p0(0) → 2
activate0(0) → 3
01() → 4
01() → 7
s1(7) → 6
n__f1(6) → 5
cons1(4, 5) → 1
s1(7) → 9
p1(9) → 8
f1(8) → 1
01() → 2
n__f1(0) → 1
f1(0) → 3
cons1(4, 5) → 3
f1(8) → 3
02() → 8
n__f2(8) → 1
n__f2(0) → 3
n__f2(8) → 3
02() → 10
02() → 13
s2(13) → 12
n__f2(12) → 11
cons2(10, 11) → 1
cons2(10, 11) → 3
0 → 3
(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(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0
Tuples:
F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
F(z0) → c2
P(s(0)) → c3
ACTIVATE(n__f(z0)) → c4(F(z0))
ACTIVATE(z0) → c5
S tuples:
F(0) → c
F(s(0)) → c1(F(p(s(0))), P(s(0)))
F(z0) → c2
P(s(0)) → c3
ACTIVATE(n__f(z0)) → c4(F(z0))
ACTIVATE(z0) → c5
K tuples:none
Defined Rule Symbols:
f, p, activate
Defined Pair Symbols:
F, P, ACTIVATE
Compound Symbols:
c, c1, c2, c3, c4, c5
(5) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
ACTIVATE(n__f(z0)) → c4(F(z0))
Removed 4 trailing nodes:
F(0) → c
P(s(0)) → c3
F(z0) → c2
ACTIVATE(z0) → c5
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0
Tuples:
F(s(0)) → c1(F(p(s(0))), P(s(0)))
S tuples:
F(s(0)) → c1(F(p(s(0))), P(s(0)))
K tuples:none
Defined Rule Symbols:
f, p, activate
Defined Pair Symbols:
F
Compound Symbols:
c1
(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
activate(n__f(z0)) → f(z0)
activate(z0) → z0
Tuples:
F(s(0)) → c1(F(p(s(0))))
S tuples:
F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:
f, p, activate
Defined Pair Symbols:
F
Compound Symbols:
c1
(9) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
activate(n__f(z0)) → f(z0)
activate(z0) → z0
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(s(0)) → 0
Tuples:
F(s(0)) → c1(F(p(s(0))))
S tuples:
F(s(0)) → c1(F(p(s(0))))
K tuples:none
Defined Rule Symbols:
p
Defined Pair Symbols:
F
Compound Symbols:
c1
(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.
F(s(0)) → c1(F(p(s(0))))
We considered the (Usable) Rules:
p(s(0)) → 0
And the Tuples:
F(s(0)) → c1(F(p(s(0))))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(F(x1)) = [3]x1
POL(c1(x1)) = x1
POL(p(x1)) = [2]
POL(s(x1)) = [3]
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:
p(s(0)) → 0
Tuples:
F(s(0)) → c1(F(p(s(0))))
S tuples:none
K tuples:
F(s(0)) → c1(F(p(s(0))))
Defined Rule Symbols:
p
Defined Pair Symbols:
F
Compound Symbols:
c1
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)