Math Problem Statement

Construct a formal proof of validity for the argument: 1. (P => Q) ∧ (R => S) 2. ¬A => ¬Q 3. A => ¬B 4. B Conclusion: ∴ ¬P ∨ ¬S

Solution

The image you've shared asks for a formal proof of validity for the given argument. Let's break down the argument and work through a proof step by step.

The Argument:

  1. (PQ)(RS)(P \Rightarrow Q) \land (R \Rightarrow S) (Conjunction of implications)
  2. ¬A¬Q\neg A \Rightarrow \neg Q (Implication)
  3. A¬BA \Rightarrow \neg B (Implication)
  4. BB (Statement)

Conclusion to Prove:

¬P¬S\therefore \neg P \lor \neg S

Step-by-Step Proof:

Step 1: Break down the conjunction from premise (1)

From the first premise, (PQ)(RS)(P \Rightarrow Q) \land (R \Rightarrow S), we can split it into two separate implications:

PQ(1a)P \Rightarrow Q \quad \text{(1a)} RS(1b)R \Rightarrow S \quad \text{(1b)}

Step 2: Derive intermediate conclusions from premises

  • From premise (4) BB, and premise (3) A¬BA \Rightarrow \neg B, we can conclude that AA must be false. If AA were true, it would lead to ¬B\neg B, which contradicts BB. Therefore:

¬A(From (3) and (4))\neg A \quad \text{(From (3) and (4))}

  • Now, use ¬A\neg A in premise (2) ¬A¬Q\neg A \Rightarrow \neg Q. Since ¬A\neg A is true, we can conclude:

\neg Q \quad \text{(From (2) and \neg A)}

Step 3: Use the contrapositive

  • We have ¬Q\neg Q. Using the contrapositive of PQP \Rightarrow Q from (1a), we know that if QQ is false, then PP must also be false:

\neg P \quad \text{(Contrapositive of (1a) using \neg Q)}

Step 4: Consider RSR \Rightarrow S

  • To derive ¬S\neg S, we turn to the second part of premise (1b), RSR \Rightarrow S. We need to argue that either RR is false or SS is false.

  • We know that the goal is to prove ¬P¬S\neg P \lor \neg S. Since we have already concluded ¬P\neg P, we do not need to further investigate whether RR or SS are true or false.

Conclusion:

We have established that ¬P\neg P. Therefore, the disjunction ¬P¬S\neg P \lor \neg S holds because one of the disjuncts is true.

Thus, the conclusion ¬P¬S\neg P \lor \neg S follows from the premises, and the argument is valid.


Would you like more details on any part of this proof? Here are 5 related questions to consider:

  1. Why do we need the contrapositive to conclude ¬P\neg P from ¬Q\neg Q?
  2. How do conjunctions like (PQ)(RS)(P \Rightarrow Q) \land (R \Rightarrow S) influence formal proofs?
  3. Can the disjunction ¬P¬S\neg P \lor \neg S still hold if ¬S\neg S were true instead of ¬P\neg P?
  4. What role does premise (3) A¬BA \Rightarrow \neg B play in leading to the conclusion?
  5. How would the proof change if the premise BB were different?

Tip: In logical proofs, applying the contrapositive is often a powerful tool when dealing with implications!

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Math Problem Analysis

Mathematical Concepts

Formal Logic
Propositional Logic
Implication
Contrapositive

Formulas

Contrapositive: (P => Q) ⇔ (¬Q => ¬P)

Theorems

Modus Ponens
Modus Tollens
Conjunction
Disjunction

Suitable Grade Level

Undergraduate