
Roman C. answered 08/25/15
Masters of Education Graduate with Mathematics Expertise
[(p → q) ∧ (q → (r ∧ s))] → (p → s)
1. Rewrite each A → B as ¬A ∨ B
¬[(¬p ∨ q) ∧ (¬q ∨ (r ∧ s))] ∨ ¬p ∨ s
2. Resolve ¬p ∨ q with ¬q ∨ (r ∧ s)
¬[¬p ∨ (r ∧ s)] ∨ ¬p ∨ s
3. DeMorgan's Law ×2
[p ∧ ¬(r ∧ s)] ∨ (¬p ∨ s)
[p ∧ (¬r ∨ ¬s)] ∨ (¬p ∨ s)
4. Distributivity of ∧ over ∨.
(p ∧ ¬r) ∨ (p ∧ ¬s) ∨ (¬p ∨ s)
5. DeMorgan's Law
(p ∧ ¬r) ∨ ¬(¬p ∨ s) ∨ (¬p ∨ s)
6. Disjunction of opposites.
(p ∧ ¬r) ∨ TRUE
7. Definition of ∨
TRUE
[(p → q) ∧ (q → (r ∧ s))] → (p → s) is therefore a tautology.