Exercise b) Recall that a formula ϕ is a tautology if and only if ¬ϕ is not satisfiable, cf. Exercise 1.2.61in Rosen that we have covered in class, that is, to check whether ϕ is a tautology, you can check whether¬ϕ is satisfiable, and reverse the answer. Note also, that since the tableau rules only deal with negation andconjunction, you have to rewrite ϕ such that it only contains these two connectives. Now, make tableaus, asthey are presented in Priest’s book, Section 1.4, showing that the formulas p → (p ∨ q) and ¬p → (p → q)are tautologies, cf. exercises 1.2.11b and 1.2.11c in Rosen’s book

To solve this problem, we need to first rewrite the formulas using only negation and conjunction, as the tableau rules only deal with these two connectives.

1. The formula p → (p ∨ q) can be rewritten as ¬p ∨ (p ∨ q), which is equivalent to ¬p ∨ p ∨ q.

2. The formula ¬p → (p → q) can be rewritten as p ∨ ¬p ∨ q, which is equivalent to p ∨ q.

Now, we can create tableaus for these formulas:

1. For the formula ¬p ∨ p ∨ q, we start with a root labeled ¬p ∨ p ∨ q. Since this is a disjunction, we create two branches: one for ¬p and one for p ∨ q. The branch for ¬p is closed because it contradicts the assumption p. The branch for p ∨ q is also closed because it contradicts the assumption ¬p. Since all branches are closed, the formula is a tautology.

2. For the formula p ∨ q, we start with a root labeled p ∨ q. Since this is a disjunction, we create two branches: one for p and one for q. Neither branch can be closed, so the formula is not a tautology.

Therefore, the formula p → (p ∨ q) is a tautology, but the formula ¬p → (p → q) is not.