(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:
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 Cpx (relative) TRS 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(cons1(z0, cons(z1, z2))) → c
2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
FROM(z0) → c2
FROM(z0) → c3
S(z0) → c4
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(z0) → c7
S tuples:
2ND(cons1(z0, cons(z1, z2))) → c
2ND(cons(z0, z1)) → c1(2ND(cons1(z0, activate(z1))), ACTIVATE(z1))
FROM(z0) → c2
FROM(z0) → c3
S(z0) → c4
ACTIVATE(n__from(z0)) → c5(FROM(activate(z0)), ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(S(activate(z0)), ACTIVATE(z0))
ACTIVATE(z0) → c7
K tuples:none
Defined Rule Symbols:
2nd, from, s, activate
Defined Pair Symbols:
2ND, FROM, S, ACTIVATE
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7
(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 5 trailing nodes:
S(z0) → c4
FROM(z0) → c2
2ND(cons1(z0, cons(z1, z2))) → c
ACTIVATE(z0) → c7
FROM(z0) → c3
(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:
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
(5) CdtRhsSimplificationProcessorProof (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)) → c1(ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
S tuples:
2ND(cons(z0, z1)) → c1(ACTIVATE(z1))
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
2nd, from, s, activate
Defined Pair Symbols:
2ND, ACTIVATE
Compound Symbols:
c1, c5, c6
(7) CdtLeafRemovalProof (ComplexityIfPolyImplication transformation)
Removed 1 leading nodes:
2ND(cons(z0, z1)) → c1(ACTIVATE(z1))
(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:
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
S tuples:
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:
2nd, from, s, activate
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c5, c6
(9) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
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
(10) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
S tuples:
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
K tuples:none
Defined Rule Symbols:none
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c5, c6
(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__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
We considered the (Usable) Rules:none
And the Tuples:
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(ACTIVATE(x1)) = [2]x1
POL(c5(x1)) = x1
POL(c6(x1)) = x1
POL(n__from(x1)) = [1] + x1
POL(n__s(x1)) = [1] + x1
(12) Obligation:
Complexity Dependency Tuples Problem
Rules:none
Tuples:
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
S tuples:none
K tuples:
ACTIVATE(n__from(z0)) → c5(ACTIVATE(z0))
ACTIVATE(n__s(z0)) → c6(ACTIVATE(z0))
Defined Rule Symbols:none
Defined Pair Symbols:
ACTIVATE
Compound Symbols:
c5, c6
(13) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty
(14) BOUNDS(1, 1)