Question

Which of the following first order formulae is logically valid? Here $\alpha \left(x\right)$ is a first order formula with x as a free variable, and $\beta$ is a first order formula with no free variable.

• A. $[\beta \to (\exists x,\alpha \left(x\right)\left)\right]\to [\forall x\beta \to \alpha (x)$]
• B. $[\exists x,\beta \to \alpha (x\left)\right]\to [\beta \to (\forall x,\alpha \left(x\right)\left)\right]$
• C. $\left[\right(\exists x,\alpha \left(x\right))\to \beta ]\to [\forall x,\alpha (x)\to \beta ]$
• D. $\left[\right(\forall x,\alpha \left(x\right))\to \beta ]\to [\forall x,\alpha (x)\to \beta ]$

Answer 1

The correct option is C $\left[\right(\exists x,\alpha \left(x\right))\to \beta ]\to [\forall x,\alpha (x)\to \beta ]$

Option (c) is $\left[\right(\exists x,\alpha \left(x\right))\to \beta ]\to [\forall x,\alpha (x)\to \beta ]$ Let us check the validity of this predicate. Let the LHS of this predicate be true.
This means that some $\alpha \to \beta$
Let ${\alpha }_5\to \beta$
Now we will check if the RHS is true. The RHS is $[\forall x,\alpha (x)\to \beta ]$ to check this implication let us take $\forall x,\alpha \left(x\right)$ to be true.
This means that all the a are true. It means that ${\alpha }_5$ is also true.
But ${\alpha }_5\to \beta$. Therefore $\beta$ is true.
So the RHS $[\forall x,\alpha (x)\to \beta$] is true. Whenever the LHS $\left[\right(\exists x,\alpha \left(x\right))\to \beta ]$ is true. So option (c) is valid.