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When I explain this to new computing science students, I approach it from the negations of these statements: ¬(B → S) ≡ B ∧ ¬S. I prove both directions of this equivalence.

Proof of (⟹): Suppose that the material conditional B → S were false. Then:

• If S, clearly, nothing can be false about the conditional: we are not talking about causation but logical implication. Hence, ¬S holds.
• If ¬B, then the conditional B → S is meaningless - how could we say it were false? Therefore, B holds.

Proof of (⟸): Suppose that B ∧ ¬S. Then B is true, but S is false. So the conditional B → S is false.

If ¬(B → S) ≡ B ∧ ¬S holds, then also B → S ≡ ¬B ∨ S. The disjunction here corresponds to the two bullet points above.

When I explain this to new computing science students, I approach it from the negations of these statements: ¬(B → S) ≡ B ∧ ¬S. I prove both directions of this equivalence.

Proof of (→): Suppose that the material conditional B → S were false. Then:

• If S, clearly, nothing can be false about the conditional - the only way it can be falsified is if S does not hold (this is logical implication, not causation). Hence, ¬S holds.
• If ¬B, then the conditional B → S is meaningless - how could we say it were false? Therefore, B holds.

Proof of (←): Suppose that B ∧ ¬S. Then B is true, but S is false. So the conditional B → S is false.

If ¬(B → S) ≡ B ∧ ¬S holds, then also B → S ≡ ¬B ∨ S. The disjunction here corresponds to the two bullet points above.

When I explain this to new computing science students, I approach it from the negations of these statements: ¬(B → S) ≡ B ∧ ¬S.

Suppose that the material conditional B → S were false. Then:

• If S, clearly, nothing can be false about the conditional: we are not talking about causation but logical implication. Hence, ¬S holds.
• If ¬B, then the conditional B → S is meaningless - how could we say it were false? Therefore, B holds.

In the other direction: suppose that B ∧ ¬S. Then the antecedent holds, but the consequent does not, so clearly the conditional B → S cannot hold.

If ¬(B → S) ≡ B ∧ ¬S holds, then also B → S ≡ ¬B ∨ S. The disjunction here corresponds to the two bullet points above.

When I explain this to new computing science students, I approach it from the negations of these statements: ¬(B → S) ≡ B ∧ ¬S. I prove both directions of this equivalence.

Proof of (⟹): Suppose that the material conditional B → S were false. Then:

• If S, clearly, nothing can be false about the conditional: we are not talking about causation but logical implication. Hence, ¬S holds.
• If ¬B, then the conditional B → S is meaningless - how could we say it were false? Therefore, B holds.

Proof of (⟸): Suppose that B ∧ ¬S. Then B is true, but S is false. So the conditional B → S is false.

If ¬(B → S) ≡ B ∧ ¬S holds, then also B → S ≡ ¬B ∨ S. The disjunction here corresponds to the two bullet points above.

When I explain this to new computing science students, I approach it from the negations of these statements: ¬(B → S) ≡ B ∧ ¬S.

Suppose that the material conditional B → S were false. Then:

• If S, clearly, nothing can be false about the conditional: we are not talking about causation but logical implication. Hence, ¬S holds.
• If ¬B, then the conditional B → S is meaningless. Therefore, B holds.

In the other direction: suppose that B ∧ ¬S. Then the antecedent holds, but the consequent does not, so clearly the conditional B → S cannot hold.

If ¬(B → S) ≡ B ∧ ¬S holds, then also B → S ≡ ¬B ∨ S. The disjunction here corresponds to the two bullet points above.

When I explain this to new computing science students, I approach it from the negations of these statements: ¬(B → S) ≡ B ∧ ¬S.

Suppose that the material conditional B → S were false. Then:

• If S, clearly, nothing can be false about the conditional: we are not talking about causation but logical implication. Hence, ¬S holds.
• If ¬B, then the conditional B → S is meaningless - how could we say it were false? Therefore, B holds.

In the other direction: suppose that B ∧ ¬S. Then the antecedent holds, but the consequent does not, so clearly the conditional B → S cannot hold.

If ¬(B → S) ≡ B ∧ ¬S holds, then also B → S ≡ ¬B ∨ S. The disjunction here corresponds to the two bullet points above.