$\phi$ satisfies $\psi_1 \cup \psi_2$ then $\phi$ satisfies $\psi_1$ or $\phi$ satisfies $\psi_2$

Question

The question was given to me in a recitation and the teacher assistant isn't available at the moment. The counter-example we were given was to define: $$\phi = p_1 \cup p_2$$ $$\psi_1 = p_1$$ $$\psi_2 = p_2$$ And then to look at $v_1$ $v_2$ where: $$v_1(\psi_1) = v_1(p_1) = t , v_1(\psi_2) = v_1(p_2) = f$$ $$v_2(\psi_1) = v_2(p_1) = f , v_2(\psi_2) = v_2(p_2) = t$$ We then get $v_2(\phi) = v_2(p_1 \cup p_2) = t$ and $v_1(\phi) = v_1(p_1 \cup p_2) = t$ therefore $\phi$ doesn't satisfy $\psi_1$ or $\psi_2$.

I would like to understand why this counter-example works because for me it looks like the two insertions still satisfy $p_1$ or $p_2$.

Answer 1

Is it true that :

if $\varphi \vDash ψ_1 \lor ψ_2$ (that means : $ψ_1 ∨ ψ_2$ is a logical consequence of $\varphi$), then either $\varphi \vDash ψ_1$ or $\varphi \vDash ψ_2$ ?

The answer is : NO.

---

As suggested, consider $\varphi := p_1 \lor p_2$ and $\psi_1 := p_1$ and $\psi_2 := p_2$.

Thus, $ψ_1 \lor ψ_2$ is $p_1 \lor p_2$ and obviously :

$p_1 \lor p_2 \vDash p_1 \lor p_2$.

But $p_1 \lor p_2 \nvDash p_1$ and $p_1 \lor p_2 \nvDash p_2$.

To show the first one, consider a truth assignment $v$ such that $v(p_1)=\text{F}$ and $v(p_2)=\text{T}$.

Clearly : $v(p_1 \lor p_2)=\text{T}$ and thus it is not true that every truth assignment that satisfies $p_1 \lor p_2$ satisfies also $p_1$.

And the same for the other case, with $v'$ such that $v'(p_2)=\text{F}$ and $v'(p_1)=\text{T}$.