The present question is about a double implication.

I have the following implications:

$(a)\implies (b)\implies ((c)$ or $(a))$

Can I deduce that $(b)\implies(a)$ only?


Although truth tables provide a systematic approach to problems like this, it's often more efficient to just think about "what can go wrong?" --- in this case, how (if at all) could the assumptions $(a\implies b)$ and $(b\implies(c\lor a))$ be true while the proposed conclusion $(b\implies a)$ is false? Falsehood of an implication is quite restrictive; it requires both that the antecedent is true and that the consequent is false, so in this case, to falsify the proposed conclusion, you'd need $b$ true and $a$ false. That automatically makes the first hypothesis, $(a\implies b)$, true. As for the second hypothesis, its antecedent $b$ is true, so the only way the implication could be true is for the consequent to be true. In this case, you'd need $(c\lor a)$ to be true. Since $a$ is false, you'd need $c$ true. In this way you find a truth assignment, "$a$ false and both $b$ and $c$ true," that makes your hypotheses both true and the proposed conclusion false, so the inference is not correct.


$(a)\implies (b)\implies ((c)$ or $(a))$

Can I deduce that $(b)\implies(a)$ only?


What you have is the following:

$$(a) \implies (b)\tag{1}$$ $$(b)\implies (c\lor a) \equiv \lnot b \lor c \lor a\tag{2}$$

In (2) I use the equivalence $p\implies q \equiv \lnot p \lor q$

From (2), It may very well have that each $\lnot b$ is true, and $a$ is true, and $c$ is true. It may very well be that we have only $\lnot b$ is true, or only $a$ is true, or only $c$ is true. Or it may very well be that we have only $\lnot b\lor c\equiv b\implies c$ is true. Or, we may have that $\lnot b \lor a \equiv b\implies a$ is true. Or both these last cases.

In short, $b\implies a$ is satisfiable (possible), but it is not logically necessary/valid.

\begin{array}{a|b|c|aVc|rem|} \hline a&b&c&a\vee c& \\ \hline F&F&F&F&\\ F&F&T&T\\ F&T&F&F\\ F&T&T&T& \text{Here, b $\rightarrow (a \lor c)$ is true, but $b \rightarrow a$ is false.}\\ T&F&F&T\\ T&F&T&T\\ T&T&F&T\\ T&T&T&T\\ \end{array}

There's no logical justification for concluding it must be the case that $b\implies a$.

**However, if you know that $a \equiv \lnot c$, then that counterexample is ruled out, because in the above indicated counterexample, we have $a = c = T$.

Now, if you also have the premise $$\lnot c\tag{3},$$ then $(3)$ together with $(2)$ imply that $\lnot b \lor a \equiv b\implies a$, by the disjunctive syllogism $\lnot b \lor c \lor a \equiv (\lnot b \lor a) \lor c$:

$\;\;\;(\lnot b \lor a) \lor c$
$\;\;\;\lnot c$
$\therefore (\lnot b \lor a)\equiv b\implies a$.

  • 1
    $\begingroup$ I'm not entirely clear what you mean by converse of (a). If you have that $\lnot c$, then you can conclude $b \implies a$ $\endgroup$ – Namaste Dec 28 '12 at 17:12
  • $\begingroup$ Yes, this is the case. Thank you very much. $\endgroup$ – Helena Dec 28 '12 at 17:14
  • $\begingroup$ Just wanted to emphasize that "It may very well be that b⟹c" is not equivalent to "b⟹c" $\endgroup$ – alancalvitti Dec 28 '12 at 17:15
  • $\begingroup$ Not a problem. $\lnot c$ together with $\lnot b \lor a \lor c$ gives you $\lnot b \lor a$ which is equivalent to $b\implies a$. $\endgroup$ – Namaste Dec 28 '12 at 17:16
  • $\begingroup$ @ amWhy: Sorry, this is an error when clicking in the page. Your solution is perfect. $\endgroup$ – Helena Feb 13 '13 at 20:49

See a truth table. It is often helpful.

$$ \begin{array}{a|b|c|aVc|rem|} \hline a&b&c&a\vee c& \\ \hline F&F&F&F&\\ F&F&T&T\\ F&T&F&F\\ F&T&T&T& \text{Here, b $\implies$a $\vee$ c is true but b $\implies$a is not.}\\ T&F&F&T\\ T&F&T&T\\ T&T&F&T\\ T&T&T&T\\ \end{array} $$ Note that False can imply Truth but truth cannot imply False.

So your statement is SATISFIABLE but not VALID.

$\textbf{Edit}$ Edited in response to some silly mistake.

  • 1
    $\begingroup$ The OP is not asking about $a \lor b$ or $b \lor a \equiv \lnot b\implies a$ Your truth table does not apply to this question. $\endgroup$ – Namaste Dec 28 '12 at 17:34
  • $\begingroup$ Still I have some trouble: Now, if (c) is the logical converse of (a). Then the implication holds. right! $\endgroup$ – Helena Dec 28 '12 at 18:00
  • $\begingroup$ Yes. If that precondition is settled, what you state is true. But your statement in OPost explicitly does not state the condition that c is converse of a. If by that you mean if a then ¬c or ¬a then c. But that is not standard language. Converse actually means something else. $\endgroup$ – 007resu Dec 28 '12 at 18:02
  • $\begingroup$ For what converse actually means see: en.wikipedia.org/wiki/Converse_(logic) $\endgroup$ – 007resu Dec 28 '12 at 18:05
  • 1
    $\begingroup$ If a = -c, then we have $b \implies (\lnot c \lor c)$, which is true no matter what the truth of $c$. So my original post still holds: concluding $b \implies a$ is satisfiable, but not valid. $\endgroup$ – Namaste Dec 28 '12 at 18:23

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