How to find this equivalent of the material conditional?

[AimaneSN]

Hi there,

It's well known that for two assertions A and B : A → B is equivalent to (nonA or B).

The only proof I know of this equivalence relies on the truth table, one just brute forces all the possible combinations of truth values and then notice they're the same every time with A → B and (nonA or B).

But how can we find the expression (nonA or B) in the first place ? I want some mechanical way that starts with A → B and gets us to (nonA or B)?

Thank you for reading.

 

[Hill]

If the definitions of the symbols via truth tables is all you got, then that is the only way.

 

[pasmith]

I have always understood "A implies B" to be defined as "it is not the case that A is true and B is false", which by Boole's laws is equivalent to "A is false or B is true".

 

[Erland]

We can also view it this way:

1. $A\rightarrow B$. (Hypothesis)
2. $A$. (Hypothesis)
3. $B$. (1, 2: Modus ponens)
4. $\neg A \lor B$. (3: Introduction of disjunction)
Thus: $A\rightarrow B, \ A\vdash\neg A \lor B$.

1. $A\rightarrow B$. (Hypothesis)
2. $\neg A$. (Hypothesis)
3. $\neg A \lor B$. (2: Introduction of disjunction)
Thus: $A\rightarrow B, \ \neg A\vdash\neg A \lor B$.

The two conclusions now give $A\rightarrow B\vdash\neg A \lor B$,
since if $A\rightarrow B$, then $\neg A \lor B$ holds whether $A$ or $\neg A$ holds.

The converse also holds:

1. $\neg A \lor B$. (Hypothesis)
2. $A$. (Hypothesis)
3. $B$. (1, 2: Elimination of disjunction)
Thus, $\neg A \lor B,\ A \ \vdash B$, and by introduction of implication: $\neg A \lor B\ \vdash A\rightarrow B$.