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Been toying with the idea of mathematizing logic, essentially finding mathematical operation analogs to logical connectives. My attempt vide infra

For 1 = True = T and 0 = False = F

a) Logical AND (p ∧ q)
T ∧ F is F and 1 × 0 = 0
F ∧ T is F and 0 × 1 = 0
F ∧ F is F and 0 × 0 = 0
T ∧ T is T and 1 × 1 = 1
Multiplication fits like a glove with conjunction

b) Logical OR (p ∨ q)
F ∨ F is F and 0 + 0 = 1
T ∨ F is T and 1 + 0 = 1
F ∨ T is T and 0 + 1 = 1
T ∨ T is T and 1 + 1 = 1 [a] Addition is almost analogous to disjunction. [a], the last line is arithmetically false. It would've worked if the operation was multiplication because 1 × 1 = 1.

c) Logical Negation (¬p)
¬T is F, which means Operation(1) = 0. The only mathematical operation that comes to mind is log 1 = 0
¬F is T, which means Operation(0) = 1. Could it be 0! (0 *factorial)?. More appropriately, we could say 10^0 = 1
Logarithms and exponentiation are converse/inverse function. So looks good, oui?

d) Logical Conditional (p → q)
T → T is T and 1 ⊕ 1 = 1 (corresponds to addition, logical disjunction AND multiplication, logical conjunction)
F → T is T and 0 ⊕ 1 = 1 (corresponds to addition, logical disjunction)
F → F is T and 0 ⊕ 0 = 1 [b]
T → F is F and 1 ⊕ 0 = 0 (corresponds to multiplication, logical conjunction)
I've used the symbol ⊕, it's a designer function or a hypothetical operation that should replicate exactly the logical conditional's processing of truth values T and F.

Also, I studied rudimentary Boolean Algebra (thousands of years ago) and much of what is in this post is from fragmented memory files.

My question is can we find (have they already been found?) mathematical designer functions/operations that faithfully mirror the logical connectives listed above insofar as what is T and F for the latter is 1 and 0 for the former?

Muchas gracias, have an awesome day!

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    I'm afraid you're a little late. This is how mathematical logic started. See the SEP entry on George Boole. plato.stanford.edu/entries/boole/#CommAlge
    – David Gudeman
    Commented Jul 6 at 5:48
  • @DavidGudeman, we could come up with our own version of logical calculus, oui? 🙂
    – Hudjefa
    Commented Jul 6 at 7:28
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    Maybe Terrence Howard should read this, especially the 1 + 1 = 1 part.
    – Hudjefa
    Commented Jul 6 at 7:41
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    You've rediscovered ring theory from the POV of generalizing arithmetic operations, and category theory from the POV of generalizing logical operations.
    – Corbin
    Commented Jul 6 at 17:56
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    @Corbin 🙂. I'm honored.
    – Hudjefa
    Commented Jul 7 at 8:58

1 Answer 1

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Thanks to David Gudeman for linking the SEP article on George Boole.

To specifically answer your question:

p∧q corresponds to pq

¬p corresponds to 1-p

You can use the above to derive formulas for the other operations.

p∨q = ¬(¬p∧¬q) corresponds to 1 - (1-p)(1-q) = p+q-pq

p→q = ¬p∨q = ¬(p∧¬q) corresponds to 1 - p(1-q) = 1 - p + pq

You can plug in 1s and 0s to verify the formulas give the correct results.

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  • I had trouble figuring out the operation for negation. It was so simple 1 - p. My log-exponential function though is unary in character, si?
    – Hudjefa
    Commented Jul 6 at 7:26
  • Does that (the above) mean anything? Negation is, last I checked, a unary operator and I would've preferred for our putative math function to mirror that: Operation(x) = 0/1, which takes only one input, viz 1 or 0.
    – Hudjefa
    Commented Jul 6 at 7:35
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    @Hudjefa, the function f(x)=1-x does just take one argument, x.
    – David Gudeman
    Commented Jul 6 at 7:54
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    @Hudjefa: That's not a question of mathematics, it's a question of notation. You can write a factorial in product notation if you really want to, the ! is just shorthand. Similarly, multiplication can be written as repeated addition, and addition as repeated succession. There's no property of the operators that makes them inherently special, we've just defined them as they are for the sake of convenience.
    – Kevin
    Commented Jul 6 at 17:26
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    @Hudjefa like Kevin said it's just notation for the same operation, but $x'$ (x prime) is a unary operator sometimes used for logical negation (or set complement) en.wikipedia.org/wiki/… as is writing a bar over the expression en.wikipedia.org/wiki/Overline#Negation
    – Pete Kirkham
    Commented Jul 6 at 19:54

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