(0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, activate(X1)))
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X
Rewrite Strategy: INNERMOST
(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted CpxTRS to CDT
(2) Obligation:
Complexity Dependency Tuples Problem
Rules:
2nd(cons1(z0, cons(z1, z2))) → z1
2nd(cons(z0, z1)) → 2nd(cons1(z0, activate(z1)))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
s(z0) → n__s(z0)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(z0) → z0
Tuples:
2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))
S tuples:
2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
2nd, from, s, activate
Defined Pair Symbols:
2ND, ACTIVATE
Compound Symbols:
c1, c5, c6
(3) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)
Split RHS of tuples not part of any SCC
(4) Obligation:
Complexity Dependency Tuples Problem
Rules:
2nd(cons1(z0, cons(z1, z2))) → z1
2nd(cons(z0, z1)) → 2nd(cons1(z0, activate(z1)))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
s(z0) → n__s(z0)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(z0) → z0
Tuples:
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))
2ND(cons(z0, z1)) → c(2ND(cons1(z0, activate(z1))))
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
S tuples:
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))
2ND(cons(z0, z1)) → c(2ND(cons1(z0, activate(z1))))
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
K tuples:none
Defined Rule Symbols:
2nd, from, s, activate
Defined Pair Symbols:
ACTIVATE, 2ND
Compound Symbols:
c5, c6, c
(5) CdtGraphRemoveTrailingProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing tuple parts
(6) Obligation:
Complexity Dependency Tuples Problem
Rules:
2nd(cons1(z0, cons(z1, z2))) → z1
2nd(cons(z0, z1)) → 2nd(cons1(z0, activate(z1)))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
s(z0) → n__s(z0)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(z0) → z0
Tuples:
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
2ND(cons(z0, z1)) → c
S tuples:
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
2ND(cons(z0, z1)) → c
K tuples:none
Defined Rule Symbols:
2nd, from, s, activate
Defined Pair Symbols:
2ND, ACTIVATE
Compound Symbols:
c, c5, c6, c
(7) CdtKnowledgeProof (EQUIVALENT transformation)
The following tuples could be moved from S to K by knowledge propagation:
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
2ND(cons(z0, z1)) → c
(8) Obligation:
Complexity Dependency Tuples Problem
Rules:
2nd(cons1(z0, cons(z1, z2))) → z1
2nd(cons(z0, z1)) → 2nd(cons1(z0, activate(z1)))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
s(z0) → n__s(z0)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(z0) → z0
Tuples:
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
2ND(cons(z0, z1)) → c
S tuples:
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
K tuples:
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
2ND(cons(z0, z1)) → c
Defined Rule Symbols:
2nd, from, s, activate
Defined Pair Symbols:
2ND, ACTIVATE
Compound Symbols:
c, c5, c6, c
(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
2ND(cons(z0, z1)) → c
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(2ND(x1)) = [5] + [4]x1
POL(ACTIVATE(x1)) = [2] + [4]x1
POL(c) = 0
POL(c(x1)) = x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(cons(x1, x2)) = [5] + x2
POL(n__from(x1)) = [2] + x1
POL(n__s(x1)) = [1] + x1
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:
2nd(cons1(z0, cons(z1, z2))) → z1
2nd(cons(z0, z1)) → 2nd(cons1(z0, activate(z1)))
from(z0) → cons(z0, n__from(n__s(z0)))
from(z0) → n__from(z0)
s(z0) → n__s(z0)
activate(n__from(z0)) → from(activate(z0))
activate(n__s(z0)) → s(activate(z0))
activate(z0) → z0
Tuples:
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
2ND(cons(z0, z1)) → c
S tuples:none
K tuples:
2ND(cons(z0, z1)) → c(ACTIVATE(z1))
2ND(cons(z0, z1)) → c
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
Defined Rule Symbols:
2nd, from, s, activate
Defined Pair Symbols:
2ND, ACTIVATE
Compound Symbols:
c, c5, c6, c
(11) SIsEmptyProof (EQUIVALENT transformation)
The set S is empty
(12) BOUNDS(O(1), O(1))