(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(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0 → n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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 4.
The compatible tree automaton used to show the Match-Boundedness (for constructor-based start-terms) is represented by:
final states : [1, 2, 3, 4, 5]
transitions:
cons0(0, 0) → 0
n__f0(0) → 0
n__s0(0) → 0
n__00() → 0
f0(0) → 1
p0(0) → 2
s0(0) → 3
00() → 4
activate0(0) → 5
n__f1(0) → 1
n__s1(0) → 3
n__01() → 4
activate1(0) → 6
f1(6) → 5
activate1(0) → 7
s1(7) → 5
01() → 5
n__f2(6) → 5
n__s2(7) → 5
n__02() → 5
f1(6) → 6
f1(6) → 7
s1(7) → 6
s1(7) → 7
01() → 6
01() → 7
n__f2(6) → 6
n__f2(6) → 7
n__s2(7) → 6
n__s2(7) → 7
n__02() → 6
n__02() → 7
02() → 8
n__02() → 11
n__s2(11) → 10
n__f2(10) → 9
cons2(8, 9) → 5
cons2(8, 9) → 6
cons2(8, 9) → 7
02() → 14
s2(14) → 13
p2(13) → 12
f2(12) → 5
f2(12) → 6
f2(12) → 7
03() → 12
n__f3(12) → 5
n__f3(12) → 6
n__f3(12) → 7
n__s3(14) → 13
n__03() → 8
n__03() → 14
03() → 15
n__03() → 18
n__s3(18) → 17
n__f3(17) → 16
cons3(15, 16) → 5
cons3(15, 16) → 6
cons3(15, 16) → 7
n__04() → 12
n__04() → 15
0 → 5
0 → 6
0 → 7
(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(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
Tuples:
F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
F(z0) → c2
P(s(0)) → c3(0')
S(z0) → c4
0' → c5
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__0) → c8(0')
ACTIVATE(z0) → c9
S tuples:
F(0) → c(0')
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
F(z0) → c2
P(s(0)) → c3(0')
S(z0) → c4
0' → c5
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__0) → c8(0')
ACTIVATE(z0) → c9
K tuples:none
Defined Rule Symbols:
f, p, s, 0, activate
Defined Pair Symbols:
F, P, S, 0', ACTIVATE
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8, c9
(5) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 8 trailing nodes:
ACTIVATE(n__0) → c8(0')
0' → c5
F(s(0)) → c1(F(p(s(0))), P(s(0)), S(0), 0')
F(0) → c(0')
P(s(0)) → c3(0')
ACTIVATE(z0) → c9
F(z0) → c2
S(z0) → c4
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
Tuples:
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c6(F(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(S(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
f, p, s, 0, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c6, c7
(7) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
Tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
f, p, s, 0, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c6, c7
(9) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
f(z0) → n__f(z0)
p(s(0)) → 0
s(z0) → n__s(z0)
0 → n__0
activate(n__f(z0)) → f(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(n__0) → 0
activate(z0) → z0
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c6, c7
(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)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(ACTIVATE(x1)) = x1
POL(c6(x1)) = x1
POL(c7(x1)) = x1
POL(n__f(x1)) = [1] + x1
POL(n__s(x1)) = [1] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
S tuples:none
K tuples:
ACTIVATE(n__f(z0)) → c6(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c7(ACTIVATE(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c6, c7
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)