If true for (x) implies true for (x + y), does this mean either always true or always false?

Question

Let $P(x)$ be some statement about real number $x.$

Suppose that $\forall x {\in} \mathbb{R} \begin{bmatrix} P(x) \implies \begin{bmatrix} \forall y {\in}\mathbb{R} \;P(x + y) \end{bmatrix} \end{bmatrix}.$

Is this sufficient to conclude that $\begin{bmatrix}\forall x {\in}\mathbb{R}\; P(x)\end{bmatrix} \text{ or } \begin{bmatrix}\forall x {\in}\mathbb{R}\;\lnot P(x)\end{bmatrix}$?

It looks sort of like induction:

We wish to show that $\begin{bmatrix}\forall x \in \mathbb{R},\, P(x) \end{bmatrix}$

First we show that $\exists x \in \mathbb{R}:\, P(x)$

Then we take $x$ to be an arbitrary element of $\mathbb{R}$

Next, we assume $P(x)$.

Then we take $y$ to be an arbitrary element of $\mathbb{R}$

Finally, we show that $P(x + y)$ is also true.

Answer 1

Suppose that $$\forall x {\in} \mathbb{R} \begin{bmatrix} P(x) \implies \begin{bmatrix} \forall y {\in}\mathbb{R} \;P(x + y) \end{bmatrix} \end{bmatrix}.$$ Is this sufficient to conclude that $$\begin{bmatrix}\forall x {\in}\mathbb{R}\; P(x)\end{bmatrix} \text{ or } \begin{bmatrix}\forall x {\in}\mathbb{R}\;\lnot P(x)\end{bmatrix}?$$

Yes, this argument is valid. In words:

For each number that satisfies P, P continues to be true however we modify that number.

Therefore, if some number satisfies P, then every number satisfies P; on the other hand, it may be that no number satisfies P.

Hence, P must either be universally true or be universally false.

Answer 2

Yes, $\forall x {\in} \mathbb{R} \left(P(x) \implies \left(\forall y {\in}\mathbb{R} \;P(x + y) \right)\right)$ implies that $\left(\forall x {\in}\mathbb{R}\; P(x)\right) \text{ or } \left(\forall x {\in}\mathbb{R}\;\lnot P(x)\right)$.

To see why, we can remark that $\forall y {\in}\mathbb{R} \;P(x + y)$ is equivalent to $\forall t {\in}\mathbb{R} \;P(t)$. Therefore, the original statement can be expressed as

$$ \forall x {\in} \mathbb{R} \left(P(x) \implies \left(\forall t {\in}\mathbb{R} \;P(t) \right)\right) $$

which implies

$$ \left(\forall x {\in}\mathbb{R}\; P(x)\right) \text{ or } \left(\forall x {\in}\mathbb{R}\;\lnot P(x)\right) $$

Either there is no real number $x$ such that $P(x)$ is true, or there is at least one, but in that case, $P(x)$ is true for every real number $x$.

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There are however problems with your reasoning:

First we show that $\exists x \in \mathbb{R}:\, P(x)$

You can't do this, since there could be no $x$ such that $P(x)$ is true.

We wish to show that $\forall x \in \mathbb{R},\, P(x)$

This is not what you want to show (because it could be false). For an example of statement $P$ where $P(x)$ is always false but $\forall x {\in} \mathbb{R} \left(P(x) \implies \left(\forall y {\in}\mathbb{R} \;P(x + y) \right)\right)$ is true, consider $P(x)$: "$x>x+1$".

Indeed, let $x$ be a real number and assume that $P(x)$ is true (that is, $x>x+1$). Then consider $y\in \mathbb R$. We have $x>x+1$ so $x+y>x+y+1$. This means that $P(x+y)$ is also true.

Answer 3

$\forall x(Px\implies\forall y:P(x+y))$

$\iff \forall x(Px\implies\forall y(True \implies P(x+y)))$

$\iff \forall x(Px\implies\forall y(\exists z:z=x+y \implies P(x+y)))$, this is true because the universe is the set of real numbers

$\iff \forall x(Px\implies\forall y\forall z(z=x+y \implies P(x+y)))$

$\iff \forall x(Px\implies\forall y\forall z(z=x+y \implies Pz))$

$\iff \forall x(Px\implies\forall y\forall z(y=z-x \implies Pz))$

$\iff \forall x(Px\implies\forall z(\exists y:y=z-x \implies Pz))$

$\iff \forall x(Px\implies\forall z(True \implies Pz))$, this is also true because the universe is the set of real numbers

$\iff \forall x(Px\implies\forall z:Pz)$

$\iff \exists x:Px\implies\forall z:Pz$

$\iff \neg\exists x:Px\lor\forall z:Pz$

$\iff \forall x:\neg Px\lor\forall z:Pz$

$\iff \forall x:\neg Px\lor\forall x:Px$

In conclusion, yes, both statements are equivalent.