July 10, 2020
Archangel Macsika
Use inference rules to deduce the following conclusions from the following sets of premises: a) Premises: p ∨ q q → r, p ∧ s → t, ¬r, ¬q → u ∧ s Conclusion: t b) Premises: (p ∧ t) → (r ∨ s), q → (u ∧ t), u → p, ¬s Conclusion: q → r
a. Solution for - Premises: p ∨ q, q → r, p ∧ s → t, ¬r, ¬q → u ∧ s Conclusion: t
| 1 | q → r | Premise |
| 2 | ¬ r | Premise |
| 3 | ¬ q | Modus Tollens of 1 and 2 |
| 4 | p ∨ q | Premise |
| 5 | p | Disjunctive syllogism of 5 and 4 |
| 6 | ¬q → u ∧ s | Premise |
| 7 | ¬ q | Copy 3 |
| 8 | u ∧ s | Modus Ponens of 7 and 6 |
| 9 | s | Simplification of 8 |
| 10 | p | Copy 5 |
| 11 | p ∧ s | Conjunction of 9 and 10 |
| 12 | p ∧ s → t | Premise |
| 13 | t | Modus Ponens of 11 and 12 |
By arriving at the conclusion "t", the argument is proven.
∴ The premise and conclusion is valid.
b. Solution for - Premises: (p ∧ t) → (r ∨ s), q → (u ∧ t), u → p, ¬s Conclusion: q → r
Since the above argument includes a proposition "q", the argument above will be valid, if the argument with the following premise and conclusion below is valid.
Knowing this, we can rewrite
Premises: (p ∧ t) → (r ∨ s)
q → (u ∧ t)
u → p
¬s
Conclusion: q → r
as
Premise: (p ∧ t) → (r ∨ s)
q → (u ∧ t)
u → p
¬s
q
Conclusion: r
Observe the two propositions and how they differ from each other. If we can prove that the second proposition is valid, it automatically means that the first (original) proposition is also valid.
| 1 | q → (u ∧ t) | Premise |
| 2 | q → u | Simplification of 1 |
| 3 | u → p | Premise |
| 4 | q → p | Hypothetical Syllogism of 2 and 3 |
| 5 | q → t | Simplication of 1 |
| 6 | q → (p ∧ t) | Conjunction of 4 and 5 |
| 7 | (p ∧ t) → (r ∨ s) | Premise |
| 8 | q → (r ∨ s) | Hypothetical Syllogism of 6 and 7 |
| 9 | q | Premise |
| 10 | r ∨ s | Modus Ponens of 8 and 9 |
| 11 | ¬s | Premise |
| 12 | r | Disjunctive syllogism of 10 and 11 |