In classical logic, does (A ⇒ C) imply ( (A∨B) ⇒ C)?

Question

does (A ⇒ C) imply ( (A∨B) ⇒ C)? it seems yes. right? this is fine. But what if A is false and B is true, will C hold? if yes can I conclude that B ⇒C?

following logic rules it seems that I can conclude B ⇒C from A ⇒C! where (A ⇒C) implies ((A∨B) ⇒ C) which implies (B ⇒C).

However, looking to the final conclusion it seems confusing that (A ⇒C) implies (B ⇒C)! can you please advise if sth went wrong?

Answer 1

When you go from

$$A \rightarrow C$$ to

$$(A \lor B) \rightarrow C$$

you are weakening the antecedent, which is not logically valid. To see why, note that the latter statement is equivalent to

$$(A \rightarrow C) \land (B \rightarrow C)$$

and so we see that weakening the antecedent ends up strengthening the statement as a whole, which is not valid: you suddenly end up claiming the additional claim of $B \rightarrow C$ whereas all you have is $A \rightarrow C$.

Concrete counterexample:

$A$: $x$ is a prime number greater than $2$

$C$: $x$ is odd

$B$: $x$ is some natural number

Now, what is valid is to strengthen the antecedent, i.e. from $A \rightarrow C$ you can infer:

$$(A \land B) \rightarrow C$$

for that ends up weakening the statement as a whole, as it is equivalent to:

$$(A \rightarrow C) \lor (B \rightarrow C)$$

Also, weakening the consequent is valid, i.e. from $A \rightarrow C$ you can infer:

$$A \rightarrow (B \lor C)$$

... but strengthening the consequent is invalid, i.e. you cannot infer:

$$A \rightarrow (B \land C)$$

for that would be equivalent to:

$$(A \rightarrow B) \land (A \rightarrow C)$$

Answer 2

Note that $A \Rightarrow C$ is equivalent to $\lnot A \lor C$.

Also $(A \lor B) \Rightarrow C$ is equivalent to $\lnot(A\lor B)\lor C$ and by De Morgan's, this is $(\lnot A \land \lnot B) \lor C$.

So we can see that these conditions are not the same: you give a counterexample yourself when $A$ is false, $B$ is true and $C$ is false.

Though it may seem intuitive that we can substitute $A$ for $A \lor B$ in this expression, this is not the case. However, "$A \Rightarrow C$ implies $(A \land B) \Rightarrow C$" is true (can you "expand" the implication to prove this?).

Answer 3

$A\implies C$ doesn't prove $B\implies C$. Neither does $A\lor B\implies C$ follow from $A\implies C$.

To see this, you just need consider what happens when $A$ is false, $B$ is true, and $C$ is false. Namely you get a result $T\implies F$ (in the case of $A\implies C$ and $A\lor B\implies C$), which is false by definition. Meanwhile $A\implies C$ gives $F\implies F$, which is true.

Thus $(A\implies C)\implies (B\implies C)$ gives $F\implies T$, in this case. Which is false. So the implication isn't valid. Similarly for the other one...