3-3) Trace all input variables through the schematic below to find X, taking care to apply the NOToperator (complement) correctly.AABBсDEFG3X= AB.C + DE.FG₁сDEFG₂G₁AB.CJustificationG₁DE FAB. C. DE F-X =AB·C + DE· F3-4) Apply DeMorgan's Theorems repeatedly to remove the complements of groups of variables. Dothis two ways: A) work from outside in; B) work from the inside out. Show all your work below.3-4A) Evaluate outside to in(side)3-4B) Evaluate inside to out(side)Stepmeget emergent ansproMed vlog (3-5whud moon how (A VENDWI aintJustificationStepThis section is optional but recommended.

The objective of the question is to apply DeMorgan's Theorems to the given Boolean expression X = AB.C + DE.F and find the value of X. DeMorgan's Theorems are used to simplify Boolean expressions and are particularly useful in digital electronics and computer science.First, let's understand DeMorgan's Theorems. There are two theorems: Theorem 1 states that the complement of a conjunction is equal to the disjunction of the complements. In other words, NOT (A AND B) = (NOT A) OR (NOT B). Theorem 2 states that the complement of a disjunction is equal to the conjunction of the complements. In other words, NOT (A OR B) = (NOT A) AND (NOT B).Now, let's apply DeMorgan's Theorems to the given expression X = AB.C + DE.F. We will do this in two ways: A) working from the outside in and B) working from the inside out.A) Working from the outside in: We start by applying DeMorgan's Theorem to the entire expression. This gives us X = NOT ((NOT (AB.C)) AND (NOT (DE.F))). Now, we apply DeMorgan's Theorem to each of the terms inside the brackets. This gives us X = NOT ((NOT (A AND B AND NOT C)) AND (NOT (D AND E AND NOT F))).B) Working from the inside out: We start by applying DeMorgan's Theorem to each of the terms inside the brackets. This gives us X = (NOT A OR NOT B OR C) AND (NOT D OR NOT E OR F). Now, we apply DeMorgan's Theorem to the entire expression. This gives us X = NOT ((A AND B AND NOT C) OR (D AND E AND NOT F)).In both cases, we end up with the same result. This shows that DeMorgan's Theorems can be applied in any order to simplify a Boolean expression.DeMorgan's Theorems were applied to the Boolean expression X = AB.C + DE.F in two ways: from the outside in and from the inside out. In both cases, the result was the same, demonstrating the flexibility of DeMorgan's Theorems in simplifying Boolean expressions.

Answered: 3-3) Trace all input variables through…