Consider the following expressions:

$(i)$ false

$(ii)$ Q

$(iii)$ true

$(iv)$ P∨Q

$(v)$ ¬Q∨P

The number of expressions given above that are logically implied by $P∧(P⇒Q)$ is ___________.

My attempt:

My doubt is regarding "true". Can we logically imply "true" from "p and p implies q" ? Per my understanding "true" and "false" are the "truth values" that can be assigned to propositions. But they are not propositions themselves.

Can you explain in formal way, please?

  • 1
    $\begingroup$ Here "true" probably refers to being a tautology. $\endgroup$ – GoodDeeds Feb 17 '16 at 19:23
  • $\begingroup$ Some versions of propositional calculus have, in addition to the usual symbols, the symbol $T$ for a universally true proposition, and upside down $T$ for a universally false proposition. $\endgroup$ – André Nicolas Feb 17 '16 at 19:30
  • $\begingroup$ In $\LaTeX$ speak, that's $\top$ and $\bot$, from \top and \bot. $\endgroup$ – Rick Decker Feb 17 '16 at 21:49

True is implied by everything. You can take it to be a/any tautology, for example $(p\lor\neg p)$. Sometimes it's a propositional constant $\top$ defined to be equivalent to a tautology.

Similarly, False implies everything. You can take it be either a shorthand for a contradiction such as $(p\land \neg p)$, or a propositional constant $\bot$ defined to be equivalent to a contradiction.


Truth ($\top$, tautology) is entailed by any proposition.

$$\begin{align}p &~\vdash~ p\wedge \top \\p\wedge \top & ~\vdash~ \top \\ \hline p & ~\vdash~ \top \end{align}$$

"If anything is true, then true is true."

Likewise falsity ($\bot$, contradiction) entails anything.

$$\begin{align}\bot &~\vdash~ p\wedge \bot\\p\wedge \bot & ~\vdash~ p\\ \hline \bot & ~\vdash~ p\end{align}$$

"If false is true, then anything is true."


Given, expression is :

$$=P\land(P\implies Q)$$

$$=P\land(P'\lor Q)$$

$$=(P\land P')\lor(P\land Q)$$

$$=F\land(P\land Q)$$

$$=(P\land Q)$$

  1. $(P\land Q)\implies F = (P\land Q)'\lor F = (P'\lor Q')\lor F=(P'\lor Q')= \text{Contingency but not Tautology}$
  2. $(P\land Q)\implies Q = (P\land Q)'\lor Q = (P'\lor Q')\lor Q=P'\lor (Q'\lor Q) =P' \lor T =T =\text{Tautology} $
  3. $(P\land Q)\implies T = (P\land Q)'\lor T = T = \text{Tautology}$
  4. $(P\land Q)\implies P\lor Q = (P\land Q)'\lor P\lor Q = (P'\lor Q')\lor P\lor Q =(P'\lor P)\lor( Q\lor Q') = T \lor T = T = \text{Tautology}$
  5. $(P\land Q)\implies Q'\lor P = (P\land Q)'\lor Q'\lor P = (P'\lor Q')\lor Q'\lor P =(P'\lor P)\lor( Q'\lor Q') = T \lor Q' = T = \text{Tautology}$

Therefore, statements $(2), (3), (4)$ and $(5)$ satisfies.

Hence, answer is $4$


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