Which one of the following options is CORRECT given three positive integers $x, y$ and $z$, and a predicate
$$P\left(x\right) = \neg \left(x=1\right)\wedge \forall y \left(\exists z\left(x=y*z\right) \Rightarrow \left(y=x\right) \vee \left(y=1\right) \right)$$
The best explanation is this...
what is the domain of y here? Is it belongs to set of the factors of a given number x or y belongs to natural number.
For 1st one irrespective of x is prime or composite, antecedent of implication is always true.Then we've to check for consequence part whether that will be T or F.
For 2nd one if y belongs to N then antecedent is always false which implies that the total implication will always be true without checking consequence part.
I think y belongs to set of factors of any given x, otherwise option B can be true???
please anyone suggest
Answer is (A).
$P\left(x\right)= (\neg \left(x=1\right)\wedge \forall y\left(\exists z\left(x=y*z\right) \implies (\left(y=x\right) \vee \left(y=1\right) \right))$
Statement: $x$ is not equal to $1$ and if there exists some $z$ for all $y$ such that product of $y$ and $z$ is $x$, then $y$ is either the number itself or $1$. This is the definition of prime numbers.
$$\exists x \forall y \forall z[\times (y, z, x) \rightarrow ((y = 1) \vee (z = 1))]$$
expresses the statement "there exists a prime number" (the number $1$ also satisfies this statement).
Note here that $\times (y, z, x)$ is equivalent to $(x = y \times z)$.
but $¬(x=1)$ removes $1$ as satisfying given number in question's formula, so the option (A) is True.
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∀y(∃z(x=y∗z) says for all y thier exist a z..
x is not independent since statement true for prime no.
if statemnt true for all x no. then c is correct.
A-> B says if A is true then B is also true.
U said left side is false then how u comment about RHS.
Defination says . iF x= Y*Z then Y=x or Z=1. then only true otherwise false.
now u take x =3 then either u take z=1 or y= 3 then only true otherwise false.
So the predicate is evaluated as
P(x) = (¬(x=1))∧(∀y(∃z(x=y*z)⇒((y=x)∨(y=1))))
P(x) being true means x ≠ 1 and
For all y if there exists a z such that x = y*z then
y must be x (i.e. z=1) or y must be 1 (i.e. z=x)
It means that x have only two factors first is 1
and second is x itself.
This predicate defines the prime number.
=> (implies)has lower precedence than ∧
$$P\left(x\right) = \left(\neg \left(x=1\right)\wedge \forall y \left(\exists z\left(x=y*z\right)\right) \Rightarrow \left(y=x\right) \vee \left(y=1\right) \right)$$
(x=1) what is the negation of this ¬(x=1)= ¬x≠1 or x ≠1.
P(x)=¬(x=1)∧∀y(∃z(x=y∗z)⇒(y=x)∨(y=1)) here a simple way to solve it is according to options A. P(x) being true means that x is a prime number. put here x=1 in LHS which is TRUE so negation of this will be FALSE and false logical and with anything will be false so LHS of implication(⇒) is false and we know that if LHS of implication is false then we don't care about RHS because F⇒ ANYTHING WILL give TRUE. so
A is right option.
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