Prove/disprove if $a\vee b \Rightarrow c$ then $a\Rightarrow c$ or $b\Rightarrow c$ and vice versa

Question

$a,b,c$ are statements, $\Rightarrow$ is a tautological consequence (not a logical implication and it's not a proposition).

Prove/disprove:

1. if $a\vee b \Rightarrow c$ then $a\Rightarrow c$ or $b\Rightarrow c$

2. if $a\Rightarrow c$ or $b\Rightarrow c$ then $a\vee b \Rightarrow c$

I'm not sure how to prove it but I think both are true, since $c$ is a tautological consequence then it's true regardless of of what $a$ and $b$ are. $c$ is true in both cases especially if $a=b=F$.

Answer 1

For 1. [using the "more standard" $\vDash$ for tautological consequence] :

if $a ∨ b \vDash c$, then $a \vDash c$ or $b \vDash c$,

consider a valuation $v$ such that one of $a,b$ is true.

We can assume, without loss of generality, that $v(a)=$t; then $v(a \lor b)=$t.

From $a ∨ b \vDash c$ it follows that $v(c)=$t, and thus : $a \vDash c$, and finally $a \vDash c$ or $b \vDash c$.

The same if we assume $v(b)=$t.

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For 2. :

if $a \vDash c$ or $b \vDash c$, then $a ∨ b \vDash c$,

it is not true, as we can see with the following counter-example.

Consider : as $a$ the formula $q \land \lnot q$, as $b$ the formula $r$ and as $c$ the formula $p \land \lnot p$.

In this case we have :

$a \vDash c$ and not $b \vDash c$,

and thus : $a \vDash c$ or $b \vDash c$, but clearly :

$(q \land \lnot q) \lor r \nvDash (p \land \lnot p)$.

Answer 2

Claim 2 is false as you can see from valuation with $v(b)=v(c)=F$ and $v(a) = T$. Claim 1 is true. It may well be near a rule of inference that $a \vee b \rightarrow c$ if and only if $(a \rightarrow c) \wedge (b \rightarrow c)$, and it's easy to show $d \wedge e \rightarrow d \vee e$. And it's tautology since no quantification rules involved.