If we consider two predicates:

$b(x)$: x is a boy

$c(x)$: x is clever

Then, there are four statements involving $∀, b(x), c(x), →$ and $↔$ . These are below along with my interpretation of their meaning. Correct me if I am wrong in any of these interpretations.

(a) $∀x(b(x)→c(x))$ - All boys are clever

(b) $∀x(b(x))→∀x(c(x))$ - If all are boys, then all are clever

(c) $∀x(b(x)↔c(x))$ - All boys are clever & all those who are clever are boys

(d) $∀x(b(x))↔∀x(c(x))$ - If all are boys, then all are clever & if all are clever then all are boys

I am trying to determine all conditional $(→,↛)$ / bi-conditional $(↔)$ relations between each two of above four. Between any two of these, if there is $↔$, then there is only one relation. However, if there is conditional relation in one direction, then conditional relation in other direction will be invalid. That is there can be $(→,↛)$. So at max there will be $4C_2\times 2=12$ such relations.

Now to understand these relations, I prepared below table. I considered that there are only two person in the universe $x_1,x_2$. $\{(b,c),(¬b,¬c)\}$ means $x_1$ is boy and is clever and $x_2$ is not a boy and is not clever. The truth values (in table) in boldface (all other than vi, viii and xiv) are one which I feel are correct, though I am not sure. The non bold-faced truth values (vi, viii and xiv) are ones which I am unsure. So, I am more likely make mistake with them.

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Based on the truth values in above table, I prepared following relations:

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Q. Are relations 1 to 12 correct?

Q. Also is there some fault in the example predicates $b(x)$ and $c(x)$ themselves?

Actually, I must doing similar for existential quantifier. But afterwards. But am I correct with all this thinking or am just over thinking? Or am severly screwed with my logics?

  • $\begingroup$ The truth values you got for your choices of model are correct - unfortunately, those four models do not give you the full picture and suggest some false relations between (a)-(d) (for example, consider what happens for $\{(b,\neg c),(\neg b,\neg c\}$), $\endgroup$ Jan 3, 2016 at 12:50
  • $\begingroup$ For $\{(b,¬c),(¬b,¬c)\}$, $(a)=false$ as $x_1$ is boy but not clever, $(b)=false$ as not all are boys, $(c)=false$ as $x_1$ is boy but not clever, $(d)=false$ as not all are boys. So all are false. That means still those ten relations do hold. In fact, I did put some thought to make choice of model so that it will lead to correct relations. Can you guess any other combination of $(b,c)$ that will make any of these ten relations fail?.... $\endgroup$
    – Mahesha999
    Jan 4, 2016 at 12:20
  • $\begingroup$ ....What I feel is these ten relations are valid for any model. For example, it will always be: $(c)→(a)$ & $(a)↛(c)$ as $c→b$ is missing from $(a)$. The real world meaning of $b$ and $c$ is irrelevant, their truth values only matter. And truth values depend on conditional & biconditional. So as long as we are dealing with these four $((a),(b),(c),(d))$ statements involving conditionals & biconditionals, these relations will hold. This is what I want confirmation for - the relation between distribution of quantifier over $→$ & $↔$. Am I wrong with these ten relations? $\endgroup$
    – Mahesha999
    Jan 4, 2016 at 12:20
  • $\begingroup$ For $\{(b,\neg c),(\neg b,\neg c)\}$, both $\forall x(b(x))$ and $\forall x(c(x))$ are false, and therefore (b) and (d) are true. In general, to check such relationships, you would need to formally check the satisfiability of their negation (using e.g. resolution), rather than relying on examples. $\endgroup$ Jan 4, 2016 at 13:22
  • $\begingroup$ my bad, said that in hurry, yes u are correct ${(b,¬c),(¬b,¬c)}{(b,¬c),(¬b,¬c)}$, $(b)=(c)=true$. So I guess my first two relations involving biconditional were wrong. It should be 1. $(a)→(b)$ 2. $(b)↛(a)$ 3. $(c)→(d)$ 4. $(d)↛(c)$ In fact my book says only $(a)→(b)$, so why I stretched myself to find these others. So does removing original first two and adding these four make the list of 12 relations correct?(If incorrect I will put more thoughts on these and also on corresponding relations with existential quantifier).... $\endgroup$
    – Mahesha999
    Jan 4, 2016 at 13:59

1 Answer 1


Reasoning based on examples is fine for disproving implications, as you have done. In order to prove that an implication holds, however, you need something more elaborate. One possibility, as mentioned in the comments, is to use resolution to prove the implications. For example, suppose you want to show that (a) implies (b), i.e. that

$\varphi:=(\forall x.(b(c)\to c(x)))\to((\forall x.b(x))\to(\forall x.c(x)))$

is valid. We will do this by deriving a contradiction from $\neg\varphi$, which (using standard equivalences) is equivalent to

$\forall x.(\neg b(x)\vee c(x)))\wedge\forall x.b(x)\wedge\exists x.\neg c(x)$.

Skolemization results in

$(\neg b(x)\vee c(x))\wedge b(x)\wedge\neg c(a)$,

from which resolution derives first $\neg b(a)$ and then an empty clause, i.e. false. Therefore the original implication is valid.

As for your table, note that the example I gave also shows that (d) does not imply (a); the remaining 5 implications look ok.

  • $\begingroup$ Hey very thanks. Wasnt aware of Skolemization. (1) Also I added 5th row in first table which indicates that $b↛a$ and $d↛c$. Thus removed first two bi-conditional relations from 2nd table & added these two non implicant. This should be correct I guess. (2) Also now I am guessing what are relations between similar statements involving existential quantifiers? (3) Just to point out what I said $b↛a$ is incorrect as it is actually "it is not the case that $b→a$" which is different from $b↛a$, as $(b↛a)≡¬(b→a)≡b∧¬a$. So relations 2,4,6,8,9,11 are incorrect. Right? $\endgroup$
    – Mahesha999
    Jan 7, 2016 at 13:46
  • $\begingroup$ Actually no, they are correct, but we must be clear about what they say: namely that (e.g.) $b\to a$ is not valid, i.e. it is not satisfied for all models. This is not the same as the negation $\neg(b\to a)$ being valid! In fact, since there are some models which satisfy $a\to b$ and some which don't, neither of them is valid. Other than that, you still have (d) implying (a) in the table; this is not the case due to the counterexample from the comments. $\endgroup$ Jan 7, 2016 at 13:58
  • $\begingroup$ With this I am not trying to find relations which are correct for this specific model, but looking for relations which are true for any set of predicates and their truth values. So apart from $d\to a$ all implication relations are valid right? $\endgroup$
    – Mahesha999
    Jan 7, 2016 at 14:42

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