Jay C. answered 3d
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I will focus on question 2.
Given A → B show that (A ∧ C) → (B ∧ C). I am using & for logical AND,
Now A → B = ¬A ∨ B. Hence ¬A ∨ B is true. This means at least one of ¬A and B is true.
Case 1. ¬A is true.
=> ¬A ∨ ¬B is true.
=> ¬(A ∧ B) is true since ¬A ∨ ¬B = ¬(A ∧ B)
=> ¬(A ∧ C) ∨ (B ∧ C) since one part of OR is true.
QED Since ¬(A ∧ C) ∨ (B ∧ C) = (A ∧ C) → (B ∧ C).
Case 2. B is true.
Case 2.1 C is true.
=> B ∧ C is true since B and C are both true.
=> ¬(A ∧ C) ∨ (B ∧ C) since one part of OR is true.
QED Since ¬(A ∧ C) ∨ (B ∧ C) = (A ∧ C) → (B ∧ C).
Case 2.1 C is false.
=> A ∧ C is false since A and C are both not true.
=> ¬(A ∧ C) is true.
=> ¬(A ∧ C) ∨ (B ∧ C) since one part of OR is true.
QED Since ¬(A ∧ C) ∨ (B ∧ C) = (A ∧ C) → (B ∧ C).
QED Since we proved the statement in each case.