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Let P(x), Q(x), R(x) and S(x) denote the following predicates with domain Z:

P(x): x ≤ 0,

Q(x): x2 = 1,

R(x): x is odd

S(x): x = x + 1.

For each predicate, determine its truth value

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Answer: Hi!, first, Z are the integer numbers, so we only will work with them.

P(x): x ≤ 0

ok, this predicate is true if x is less or equal tan 0, and false if x is greater than 0.

so P(x) is true if { x∈Z, x ≤ 0}

Q(x): x2 = 1

Q(x) is true only if 2*x = 1. now, this means that if x=1/2 is true, but 1/2 isnt an integer, then Q(x) is false ∀ x ∈ Z.

R(x): x is odd

R(x) is true if x is odd, we can write odd numbers as x = 2k + 1, where k is a random integer; then:

R(x) is true if x=2k +1, with k∈Z.

S(x): x = x + 1

S(x) is true if x= x+1, if we subtract x from both sides of the equality, we get that S(x) is true if 0=1, and this is absurd, then:

S(x) is false ∀ x ∈ Z.

User Dheeraj Bhaskar
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