Using only quantifications, parentheses, logical connectives, variables, and the predicates d(t, x) and o(t) and q(t, x) defined in part (A), write something in place of . . . in the following to make a true statement. ∀x ∈ Z+ [¬p(x) ↔ . . . ].
Question
Using only quantifications, parentheses, logical connectives, variables, and the predicates d(t, x) and o(t) and q(t, x) defined in part (A), write something in place of . . . in the following to make a true statement. ∀x ∈ Z+ [¬p(x) ↔ . . . ].
Solution
∀x ∈ Z+ [¬p(x) ↔ (∃t ∈ Z+ [(d(t, x) ∧ ¬o(t) ∧ ¬q(t, x))])].
Here's the step-by-step explanation:
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∀x ∈ Z+ [¬p(x) ↔ . . . ]: This statement says "for all positive integers x, x is not prime if and only if . . .". We need to fill in the blank with a condition that is true when x is not a prime number.
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Prime numbers are defined as numbers that have exactly two distinct positive divisors: 1 and the number itself. So, a number is not prime if it has a divisor other than 1 and itself.
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We can express this condition using the predicates d(t, x), o(t), and q(t, x). The predicate d(t, x) means "t divides x", o(t) means "t equals 1", and q(t, x) means "t equals x".
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So, a number x is not prime if there exists a number t such that t divides x, t is not equal to 1, and t is not equal to x. We can write this condition as ∃t ∈ Z+ [(d(t, x) ∧ ¬o(t) ∧ ¬q(t, x))].
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Therefore, the complete statement is ∀x ∈ Z+ [¬p(x) ↔ ∃t ∈ Z+ [(d(t, x) ∧ ¬o(t) ∧ ¬q(t, x))]]. This statement is true for all positive integers x. It says "for all positive integers x, x is not prime if and only if there exists a positive integer t such that t divides x, t is not equal to 1, and t is not equal to x".
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