Solution to Let P(x,y) denote the sentence x2 + 1≥ x + 1. What are the truth … - Sikademy
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Archangel Macsika

Let P(x,y) denote the sentence x2 + 1≥ x + 1. What are the truth value of the following where the domain of x and y is the set of all integers? a. ⱯxⱯyP(x,y) b. ⱯxƎyP(x,y) c. ƎxⱯyP(x,y) . d. ƎxƎyP(x,y)

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As i understand, there is a mistake in condition, and P(x,y) denote the sentence x^2+ 1≥ y + 1 , otherwise it is independent from y, which doesn't make much sence.

a. ⱯxⱯyP(x,y). For x = 0 and y = 1 we have 0^2+1≥1+1\implies 1≥2 , which is false

This statement is false


b. ⱯxƎyP(x,y). Let x=a, a\isin Z , then a^2+1≥y+1\implies y≤a^2 , which means we can find such y, for exmaple, y=a^2\in Z , that P(x, y) is true

This statement is true


c. ƎxⱯyP(x,y). Let x=a, a\isin Z , then a^2+1≥y+1\implies y≤a^2, but we can put, for example,

y=a^2+1 , which means P(x, y) would be false

So, this statemnet is false


d. ƎxƎyP(x,y). For x = 0 and y = 0 we have 0^2+1=0+1\implies 1=1 . P(x, y) is true

This statement is true. Also this statement implies from the true statement (b)


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