Textbook is Ben-Ari's Mathematical Logic for Computer Science. This question is taken directly from the homework that my professor assigned, not from the textbook. Definitions of interpretations and Sub are from the textbook.

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I'm just lost at how to tackle this. First of all I don't really understand what the question is saying, especially the A. 2) part with the I'.

My attempt which is probably wrong:

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1 Answer 1


It is useful to consider some simple cases.

Consider for simplicity the case without clause (2), i.e. a formula like :

$\mathcal A := \lnot p_1$

and consider a formula $\mathcal B_1$.

Then :

$\mathcal A' := \lnot \mathcal B_1$.

We have an interpretation $I$ such that $I(p_1)=T$ and consider a new interpretation $I'$ such that $v_{I'}(\mathcal B_1)=T$.

You have to note that $\mathcal B_1$ is a formula built-up with connectives and some propositional letters, in general different from $p_1$; let $\mathcal B_1$ be $(q_1 \land ((\lnot q_2) \lor q_3))$.

$I'$ must be defined on $q_1,q_2,q_3$ and not necessarily on $p_1$; but this does not matter. The fact is that $I'$ "induces" (by the truth-table "calculations") a value for $\mathcal B_1$ and we know that this value is the same that $I$ assign to $p_1$, i.e. (in our example) $v_{I'}(\mathcal B_1)=I(p_1)=T$.

What we want to prove is that :

$v_{I}(\mathcal A)=v_{I'}(\mathcal A')$.

And this holds because we have :

$v_{I}(\mathcal A)=v_{I}(\lnot p_1)=F$

and :

$v_{I'}(\mathcal A')=v_{I'}(\lnot \mathcal B_1)=F$.

The next step is to "test the mechanism" on a slightly more complex formula like $\mathcal A := p_1 \land p_2$, where $\mathcal A' := \mathcal B_1 \land \mathcal B_2$.

Having completed this "warm-up", we have to prove it by induction on the "complexity" of $\mathcal A$, i.e. by induction on the number $n$ of occurrences of connectives in $\mathcal A$, where $\mathcal A$ is a formula built-up with propositional letters $\{ p_1, \ldots, p_k \}$.

The base case for $n=0$ is trivial; we have that $\mathcal A := p_1$ and $\mathcal A' := \mathcal B_1$.

The induction step must be performed for all the connectives of the language : $\lnot, \land, \lor$.

We have that $\mathcal A := \lnot \mathcal A_1$ or $\mathcal A := \mathcal A_1 \land \mathcal A_2$, and so on.

We assume as induction hypotheses that the property holds for the $\mathcal A_i$'s [which, of course, being subformulae of $\mathcal A$, have less occurrences of connectives than $\mathcal A$] and we use the "mechanism" above to conclude that it holds also for $\mathcal A$.

The successive step is to consider also the other propositional letters $q \notin \{ p_1, \ldots, p_k \}$.

  • $\begingroup$ Thank you Mauro as always, you've been very helpful with my logic questions $\endgroup$
    – Keith Yong
    Sep 27, 2014 at 23:18

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