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Let Δ, ∇ ∈ {∧, ∨} be such that (p → q) Δ (p ∨ q) is a tautology. Then (1) Δ = ∨, ∇ = ∨ (2) Δ = ∨, ∇ = ∧ (3) Δ = ∧, ∇ = ∨ (4) Δ = ∧, ∇ = ∧

1. Δ = ∨, ∇ = ∨
2. Δ = ∨, ∇ = ∧
3. Δ = ∧, ∇ = ∨
4. Δ = ∧, ∇ = ∧

Correct answer: Δ = ∨, ∇ = ∧

Solution

The correct option is right because using the logical operations of disjunction (∨) and conjunction (∧) in the expression (p → q) ∨ (p ∨ q) results in a tautology, meaning it is true for all truth values of p and q. This combination ensures that at least one part of the expression is always true, fulfilling the condition of being a tautology.

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