Question

Show that s is a valid conclusion from the premises. p → q, p → r, ¬(q ∧ r), s ∨ p​

Answer 1

Concept Introduction: Sets are basic mathematics.

Given:

We have been Given:

$p → q \\ p → r \\ (q ∧ r) \\ s ∨ p$

To Find:

We have to Find: Show that s is a valid conclusion from the premises.

Solution:

According to the problem,

$p → q \\ p → r \\ (q ∧ r) \\ s ∨ p$

Final Answer: The answer is

$p → q \\ p → r \\ (q ∧ r) \\ s ∨ p$

#SPJ1

Answer 2

Answer:

Step-by-step explanation:

To show that "s" is a valid conclusion from the premises "p → q", "p → r", "¬(q ∧ r)", and "s ∨ p", we can use the method of deduction.

We start by assuming the negation of the conclusion, which is ¬s.

Since "s ∨ p" is true, we can consider two cases:

a. If "s" is true, then we have already shown that the conclusion is true, so we can stop here.

b. If "s" is false, then "p" must be true in order for "s ∨ p" to be true.

From "p → q" and "p", we can use the rule of modus ponens to infer that "q" is true.

Similarly, from "p → r" and "p", we can use the rule of modus ponens to infer that "r" is true.

However, we also know that "¬(q ∧ r)" is true, which means that both "q" and "r" cannot be true at the same time.

Since we have already shown that "q" and "r" are both true, this leads to a contradiction.

Therefore, our initial assumption of ¬s must be false, and we can conclude that "s" is true.

Thus, we have shown that "s" is a valid conclusion from the given premises.