From Example 24c, we know that (4x)[P(x) ` Q(x)] 4
Chapter 1, Problem 37(choose chapter or problem)
From Example 24c, we know that (4x)[P(x) ` Q(x)] 4 (4x)P(x) ` (4x)Q(x) is valid. From Practice 21, we know that (4x)[P(x) ~ Q(x)] 4 (4x)P(x) ~ (4x)Q(x) is not valid. From Exercise 34a, we know that (E x)[P(x) ` Q(x)] 4 (E x)P(x) ` (E x)Q(x) is not valid. Explain why (E x)[P(x) ~ Q(x)] 4 (E x)P(x) ~ (E x)Q(x) is valid.
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