I How to find this equivalent of the material conditional?

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The discussion focuses on finding a mechanical way to derive the expression (¬A ∨ B) from the implication A → B. It highlights that the equivalence can be understood through logical definitions, specifically that A → B is defined as "it is not the case that A is true and B is false." The conversation presents a structured proof using hypotheses and logical rules, demonstrating how to arrive at (¬A ∨ B) from A → B through both direct and converse reasoning. The participants emphasize that if A → B holds, then (¬A ∨ B) must also hold, establishing their equivalence. The thread concludes that understanding these logical relationships is essential for deriving implications in propositional logic.
AimaneSN
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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.
 
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If the definitions of the symbols via truth tables is all you got, then that is the only way.
 
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".
 
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##.
 
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