# When is substituting the consequent of a material conditional allowed?

If $$A \implies C$$ is true, then does this imply that either of the following two sentences are true?

$$A \land B \equiv C \land B$$

$$A \lor B \equiv C \lor B$$

Intuitively, I would guess that the first sentence is true since if the first half is true, then $$A$$ and $$B$$ is true, and if $$A$$ is true, $$C$$ must also be true so $$A$$ and $$C$$ must be true. I would guess that the second sentence is false since $$B$$ and $$C$$ could be false but $$A$$ could be true, making the LHS true and the RHS false.

If my intuitions are right, would that mean that you can "substitute" the consequence of a material conditional into a sentence sometimes, but not always. If so, what are the rules for doing this?

• For the first, you explain why $\implies$ holds but you ignore the other implication. What you say about the second sentence, "B and C could be false but A could be true" is impossible since $A\implies C$ is true. Sep 18, 2021 at 21:37
• We have "$0 \implies 1$" true. With "B true" is the equivalence false. Similarly, the same choice of A,C but "B false" gives non-equivalence. Sep 18, 2021 at 21:39
• I think you would do better to draw up some truth tables rather than rely on your intuitions. Just considering the cases where $A$ is false will reveal some problems with your thinking. Sep 18, 2021 at 22:01
• next question in probability: can we just plug in the conditional like $E[XY|X=5] = E[5Y|X=5]$ ?
– BCLC
Sep 19, 2021 at 17:57

Assume $$A$$ is false and $$C$$ is true.
Then $$A \implies C$$ is true.
If B is true then $$A \land B \equiv C \land B$$ is false because LHS is false and RHS is true.
If B is false then$$A \lor B \equiv C \lor B$$ is false because LHS is false but RHS is true. Thus none of the above equivalents is true.
• $$A → C \;\not\vDash\; (A ∧ B) \leftrightarrow (C ∧ B)$$ counter-interpretation: $$(A,B,C)=(0,1,1)$$
• $$A → C \;\not\vDash\; (A ∨ B) \leftrightarrow (C ∨ B)$$ counter-interpretation: $$(A,B,C)=(0,0,1)$$