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In forall x: Calgary, by P. D. Magnus, p. 153, there is this question:

B. Formulate strategies for working backward and forward from $\mathcal{A} \leftrightarrow \mathcal{B}$.

Working forward

Having $\mathcal{A} \leftrightarrow \mathcal{B}$ as a premise, I would want to eliminate the biconditional. So, I would need to find A and justify it using a conditional. Then, I could use $\leftrightarrow E$ to get $\mathcal{B}$.

I need to use Elimination Rules when working forward and Introduction Rules, backward.

Is something similar to this what the book asks ? Not sure if he is referring to working forward from premises or $\mathcal{A} \leftrightarrow \mathcal{B}$ is a sentence that appear somewhere in the proof (or subproof).

F. Zer
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1 Answers1

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Is something similar to this what the book asks ? Not sure if he is referring to working forward from premises or $\mathcal{A} \leftrightarrow \mathcal{B}$ is a sentence that appear somewhere in the proof (or subproof).

Yes, that is it. You want strategies that take $A\leftrightarrow B$ as one of the premises, assumptions, or derivations and work forward towards deriving some target.

The rules of biconditional elimination and conditional elimination, along with having one of the equivalents ($A, B$), will build such a strategy.

Likewise you want strategies where $A\leftrightarrow B$ is the target, and work backward to derive that target.

Graham Kemp
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  • Thank you, Graham. In case I have $A\leftrightarrow B$ as one of the premises, the only thing I need to find is $A$ or $B$. Then, I can use biconditional elimination. Is it correct ? – F. Zer Mar 03 '20 at 23:59
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    @F.Zer What does the rule of biconditional elimation say? There's often some symbol like "$\vdash$" or a set of dots, or a line, or a phrase like "we may infer", which separates what you start with before using a rule and what you get when you use that rule. In the statement of such rules, the variables stand for any well-formed formulas. Thus, you can use biconditional elimination anytime you have a well-formed biconditional formula. – Doug Spoonwood Mar 04 '20 at 14:06