Formal logic question - why is this invalid?

Question

I typed this into a logic calculator, and it comes out as invalid:

\begin{align*} \forall x, & E(x) \implies I(x) \vee A(x) \\ \forall x, & A(x) \implies \neg E(x) \\ \forall x, & E(x) \implies \neg A(x)\\ \exists x, & E(x) \implies I(x)\\ \exists x, & E(x)\\ E(x)⟹I(x) \end{align*}

Can someone help me explain why it's invalid?

https://www.umsu.de/trees/#((((((%E2%88%80y((Ex%E2%86%92(Ix%E2%88%A8Ax))))%E2%88%A7(%E2%88%80y((Ax%E2%86%92(%C2%ACEx)))))%E2%88%A7(%E2%88%80y((Ex%E2%86%92(%C2%ACAx)))))%E2%88%A7(%C2%AC(%E2%88%80y((%C2%AC(Ex%E2%86%92Ix))))))%E2%88%A7(%C2%AC(%E2%88%80y((%C2%ACEx)))))%E2%88%A7(Ex%E2%86%92Ix))

Answer 1

Long comment

As you have written it in the logic calculator, it is a single formula (a conjunction) and not an argument.

Thus, the calculator's answer is correct: the formula is invalid.

As answered by the tool, consider a domain $D = \{ 0 \}$ where we have that $E(0), I(0)$ and $A(0)$ hold. The part $\forall x (Ex \to \lnot Ax)$ is not satisfied.

Answer 2

The third line is the contrapositive of the second line. The first and third lines together imply $$\forall x,E(x)\implies (I(x)\lor A(x))\land \lnot A(x)\implies I(x)$$ which is the conclusion. The fourth and fifth lines are unnecessary, but they don't contradict anything in the first three lines, so the argument seems perfectly valid to me.

I don't know anything about the calculator.

Answer 3

Mauro is right. If you enter this into the logic calculator at somerby.net/mack/logic

x,Ex->Ix|Ax
x,Ax->~Ex
x,Ex->Ix
3x,Ex->Ix
3x,Ex
Ex->Ix

it will interpret it as a conjunction of all of the lines, i.e. "Line 1 AND Line 2 AND Line 3 AND ...". What you need is the "therefore" symbol. If you enter

x,Ex->Ix|Ax
x,Ax->~Ex
x,Ex->Ix
3x,Ex->Ix
3x,Ex
,',
Ex->Ix

it will tell you that it is a valid argument, as you expect. The ",'," means "therefore". The lines before it are interpreted as the premises of the argument and the line after it are interpreted as the conclusion.