Whether following propositional formula a tautology?

$(p\iff q) \space \lor \space (p\iff \lnot q)$.

I think it is not. For contradiction I came up with following interpretation.

$p(x): \text{x got A grade in math.}$

$q(x): \text{x got A grade in english.}$

Now, in a class where there are students getting $A$ in both english and math as well as there are students getting $A$ in only one of them. Then above formula fails.

But answer given is that formula is a tautology.

They gives following reasoning for that.

As we know, $\lnot(p\iff q) = (p\iff \lnot q)$


$X \lor \lnot X$ is tautology.

So is $(p\iff q) \lor \lnot (p\iff q)$.

So, where I am getting wrong.


3 Answers 3


The formula in question does not contain quantifiers. It is a truth-functional statements, not a quantificational one. So, you cannot have one world where $p$ is true 'for some' but false 'for others' ... there are no 'some' or 'others' here. Instead, for any one world, $p$ will be one of true or false, period.

  • $\begingroup$ Can you please elaborate a little more. $\endgroup$ Nov 14, 2019 at 3:59
  • $\begingroup$ @VimalPatel I just did ... $\endgroup$
    – Bram28
    Nov 14, 2019 at 4:00

The point is that your example with grades is about the formula $$ (\forall x, p(x) \iff q(x)) \lor (\forall x, p(x)\iff \lnot q(x))$$ but actually you should be looking at $$ \forall x, \left( (p(x) \iff q(x)) \lor (p(x)\iff \lnot q(x))\right).$$


Reference for the method I apply here : Mendelson, Schaum's outline of boolean algebra and switching circuits.

In order an OR statement to be false, it is required each member of the disjunction to be false.

Supposse there is a truth value assignment such that your formula is false.

In this assignment, both members are false, so, in particular, the first is false. so we have : ( P <--> Q) is false.

This can happen in two cases : P true and Q false, P false and Q true.

Case 1 : P is true, Q is false; therefore ~ Q is true ; and finally ( P <--> ~Q) is true ( see the truth table of <--> ) ; hence a contradiction, since we initially assumed both members of the disjunction to ba false.

Case 2 : P false , Q true ; therefore ~ Q is false; and finally (P <--> ~Q) is true ; hence again a contradiction for the same reason.

Conclusion : there is no possible truth value assignment in which the formula is false; this formula is a tautology.


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