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

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

and(x, false) → false
and(x, not(false)) → x
not(not(x)) → x
implies(false, y) → not(false)
implies(x, false) → not(x)
implies(not(x), not(y)) → implies(y, and(x, y))

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(z0, false) → false
and(z0, not(false)) → z0
not(not(z0)) → z0
implies(false, z0) → not(false)
implies(z0, false) → not(z0)
implies(not(z0), not(z1)) → implies(z1, and(z0, z1))

Tuples:

IMPLIES(false, z0) → c3(NOT(false))
IMPLIES(z0, false) → c4(NOT(z0))
IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))

S tuples:

IMPLIES(false, z0) → c3(NOT(false))
IMPLIES(z0, false) → c4(NOT(z0))
IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))

K tuples:none
Defined Rule Symbols:

and, not, implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c3, c4, c5

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

Removed 2 trailing nodes:

IMPLIES(false, z0) → c3(NOT(false))
IMPLIES(z0, false) → c4(NOT(z0))

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(z0, false) → false
and(z0, not(false)) → z0
not(not(z0)) → z0
implies(false, z0) → not(false)
implies(z0, false) → not(z0)
implies(not(z0), not(z1)) → implies(z1, and(z0, z1))

Tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))

S tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))

K tuples:none
Defined Rule Symbols:

and, not, implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c5

(5) 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.

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))

We considered the (Usable) Rules:

and(z0, false) → false

And the Tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))

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

POL(AND(x₁, x₂)) = [5] + x₂   
POL(IMPLIES(x₁, x₂)) = [5]x₁ + [2]x₂   
POL(and(x₁, x₂)) = [5] + [4]x₁ + x₂   
POL(c5(x₁, x₂)) = x₁ + x₂   
POL(false) = 0   
POL(not(x₁)) = [5] + [4]x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

and(z0, false) → false
and(z0, not(false)) → z0
not(not(z0)) → z0
implies(false, z0) → not(false)
implies(z0, false) → not(z0)
implies(not(z0), not(z1)) → implies(z1, and(z0, z1))

Tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))

S tuples:none
K tuples:

IMPLIES(not(z0), not(z1)) → c5(IMPLIES(z1, and(z0, z1)), AND(z0, z1))

Defined Rule Symbols:

and, not, implies

Defined Pair Symbols:

IMPLIES

Compound Symbols:

c5

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))