(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:
from(X) → cons(X, n__from(s(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
activate(n__from(X)) → from(X)
activate(X) → X
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
FROM(z0) → c
FROM(z0) → c1
SEL(0, cons(z0, z1)) → c2
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c4(FROM(z0))
ACTIVATE(z0) → c5
S tuples:
FROM(z0) → c
FROM(z0) → c1
SEL(0, cons(z0, z1)) → c2
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c4(FROM(z0))
ACTIVATE(z0) → c5
K tuples:none
Defined Rule Symbols:
from, sel, activate
Defined Pair Symbols:
FROM, SEL, ACTIVATE
Compound Symbols:
c, c1, c2, c3, c4, c5
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 5 trailing nodes:
ACTIVATE(n__from(z0)) → c4(FROM(z0))
SEL(0, cons(z0, z1)) → c2
FROM(z0) → c1
ACTIVATE(z0) → c5
FROM(z0) → c
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2))
S tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)), ACTIVATE(z2))
K tuples:none
Defined Rule Symbols:
from, sel, activate
Defined Pair Symbols:
SEL
Compound Symbols:
c3
(5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(z0) → z0
Tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
S tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
K tuples:none
Defined Rule Symbols:
from, sel, activate
Defined Pair Symbols:
SEL
Compound Symbols:
c3
(7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
activate(n__from(z0)) → from(z0)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
Tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
S tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
K tuples:none
Defined Rule Symbols:
activate, from
Defined Pair Symbols:
SEL
Compound Symbols:
c3
(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.
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
We considered the (Usable) Rules:none
And the Tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(SEL(x1, x2)) = x1
POL(activate(x1)) = 0
POL(c3(x1)) = x1
POL(cons(x1, x2)) = 0
POL(from(x1)) = 0
POL(n__from(x1)) = 0
POL(s(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
activate(n__from(z0)) → from(z0)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
Tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
S tuples:none
K tuples:
SEL(s(z0), cons(z1, z2)) → c3(SEL(z0, activate(z2)))
Defined Rule Symbols:
activate, from
Defined Pair Symbols:
SEL
Compound Symbols:
c3
(11) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(12) BOUNDS(1, 1)