The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

(0) Obligation:

The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).


The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
sel(0, cons(X, Z)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
from(X) → n__from(X)
first(X1, X2) → n__first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__first(X1, X2)) → first(X1, X2)
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)
first(0, z0) → nil
first(s(z0), cons(z1, z2)) → cons(z1, n__first(z0, activate(z2)))
first(z0, z1) → n__first(z0, z1)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(n__first(z0, z1)) → first(z0, z1)
activate(z0) → z0

Tuples:

FROM(z0) → c
FROM(z0) → c1
FIRST(0, z0) → c2
FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
FIRST(z0, z1) → c4
SEL(0, cons(z0, z1)) → c5
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c7(FROM(z0))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))
ACTIVATE(z0) → c9

S tuples:

FROM(z0) → c
FROM(z0) → c1
FIRST(0, z0) → c2
FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
FIRST(z0, z1) → c4
SEL(0, cons(z0, z1)) → c5
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__from(z0)) → c7(FROM(z0))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))
ACTIVATE(z0) → c9

K tuples:none
Defined Rule Symbols:

from, first, sel, activate

Defined Pair Symbols:

FROM, FIRST, SEL, ACTIVATE

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 7 trailing nodes:

FROM(z0) → c
ACTIVATE(z0) → c9
FROM(z0) → c1
SEL(0, cons(z0, z1)) → c5
ACTIVATE(n__from(z0)) → c7(FROM(z0))
FIRST(z0, z1) → c4
FIRST(0, z0) → c2

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
first(0, z0) → nil
first(s(z0), cons(z1, z2)) → cons(z1, n__first(z0, activate(z2)))
first(z0, z1) → n__first(z0, z1)
sel(0, cons(z0, z1)) → z0
sel(s(z0), cons(z1, z2)) → sel(z0, activate(z2))
activate(n__from(z0)) → from(z0)
activate(n__first(z0, z1)) → first(z0, z1)
activate(z0) → z0

Tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

S tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

K tuples:none
Defined Rule Symbols:

from, first, sel, activate

Defined Pair Symbols:

FIRST, SEL, ACTIVATE

Compound Symbols:

c3, c6, c8

(5) 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))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

activate(n__from(z0)) → from(z0)
activate(n__first(z0, z1)) → first(z0, z1)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
first(0, z0) → nil
first(s(z0), cons(z1, z2)) → cons(z1, n__first(z0, activate(z2)))
first(z0, z1) → n__first(z0, z1)

Tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

S tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

K tuples:none
Defined Rule Symbols:

activate, from, first

Defined Pair Symbols:

FIRST, SEL, ACTIVATE

Compound Symbols:

c3, c6, c8

(7) 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)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))

We considered the (Usable) Rules:none
And the Tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ACTIVATE(x₁)) = 0   
POL(FIRST(x₁, x₂)) = 0   
POL(SEL(x₁, x₂)) = [2]x₁   
POL(activate(x₁)) = 0   
POL(c3(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c8(x₁)) = x₁   
POL(cons(x₁, x₂)) = 0   
POL(first(x₁, x₂)) = 0   
POL(from(x₁)) = 0   
POL(n__first(x₁, x₂)) = 0   
POL(n__from(x₁)) = 0   
POL(nil) = 0   
POL(s(x₁)) = [2] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

activate(n__from(z0)) → from(z0)
activate(n__first(z0, z1)) → first(z0, z1)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
first(0, z0) → nil
first(s(z0), cons(z1, z2)) → cons(z1, n__first(z0, activate(z2)))
first(z0, z1) → n__first(z0, z1)

Tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

S tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

K tuples:

SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))

Defined Rule Symbols:

activate, from, first

Defined Pair Symbols:

FIRST, SEL, ACTIVATE

Compound Symbols:

c3, c6, c8

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

We considered the (Usable) Rules:

first(z0, z1) → n__first(z0, z1)
from(z0) → cons(z0, n__from(s(z0)))
first(s(z0), cons(z1, z2)) → cons(z1, n__first(z0, activate(z2)))
first(0, z0) → nil
from(z0) → n__from(z0)
activate(n__from(z0)) → from(z0)
activate(z0) → z0
activate(n__first(z0, z1)) → first(z0, z1)

And the Tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(ACTIVATE(x₁)) = x₁   
POL(FIRST(x₁, x₂)) = x₂   
POL(SEL(x₁, x₂)) = x₁·x₂   
POL(activate(x₁)) = x₁   
POL(c3(x₁)) = x₁   
POL(c6(x₁, x₂)) = x₁ + x₂   
POL(c8(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₂   
POL(first(x₁, x₂)) = [1] + x₁ + x₂   
POL(from(x₁)) = 0   
POL(n__first(x₁, x₂)) = [1] + x₁ + x₂   
POL(n__from(x₁)) = 0   
POL(nil) = [1]   
POL(s(x₁)) = [2] + x₁   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

activate(n__from(z0)) → from(z0)
activate(n__first(z0, z1)) → first(z0, z1)
activate(z0) → z0
from(z0) → cons(z0, n__from(s(z0)))
from(z0) → n__from(z0)
first(0, z0) → nil
first(s(z0), cons(z1, z2)) → cons(z1, n__first(z0, activate(z2)))
first(z0, z1) → n__first(z0, z1)

Tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

S tuples:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))

K tuples:

SEL(s(z0), cons(z1, z2)) → c6(SEL(z0, activate(z2)), ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

Defined Rule Symbols:

activate, from, first

Defined Pair Symbols:

FIRST, SEL, ACTIVATE

Compound Symbols:

c3, c6, c8

(11) CdtKnowledgeProof (EQUIVALENT transformation)

The following tuples could be moved from S to K by knowledge propagation:

FIRST(s(z0), cons(z1, z2)) → c3(ACTIVATE(z2))
ACTIVATE(n__first(z0, z1)) → c8(FIRST(z0, z1))

Now S is empty

(12) BOUNDS(1, 1)