[ivan.moony]

If also has some deeper insights. For example, "if X is bird then X can fly" can be written as:

Bird(x) -> Fly(x)

Which translates to:

~Bird(x) | Fly(x)

Now, conveniently if we say "Bird(x)", resolution rule automatically asserts "Fly(x)". But also, less obvious, if we say "~Fly(x)", resolution rule asserts "~Bird(x)", which means "if X can't fly, then X is not a bird".

[8pla.net]

learn if there is light then day else night

learn if there is light then it is day else it is night

learn if (there = light) then (it = day) else (it = night)

learn[] = there = light ? it = day : it = night ;

PHP:
$light=rand(0,1);$learn[]=$there=$light?$it="day":$it="night";

The choice is decided at random.  Locate the middle equals operator in the color coded source. Should logic set (=), or compare (==), $there to $light?

[ivan.moony]

That's a tricky question. In logic, there's no distinction between getters and setters. Refer to documentation about logic for more details.

[8pla.net]

Thank you Ivan.

I am starting by converting the simplest logic to source code.
When it works with the simplest logic as source code, then I think
there is a point in building upon that foundation.

Per your request...

Code

<script language="JavaScript">
function learn(there) {
  var light=Math.floor(Math.random() * Math.floor(2));
  return (there=light ? "day" : "night");
}
var there;
console.log(learn(there));
</script>

Console log displays "day" or "night" at random.

[LOCKSUIT]

Is the material really saying to assign truths to things if its parts are true? Really? How does something's parts being true make the thing they make true?

Edit: ex. "I can eat and sleep."

[ivan.moony]

Lock,

Not in all cases. If we have a sequence:

A1 | A2 | A3 | ...

then only one element has to be true for the whole sequence to be true. On the other hand, if we have:

A1 & A2 & a3 & ...

then all the elements have to be true for the whole sequence to be true.

[LOCKSUIT]

Yes, I got that about the OR :p
Including how in some cases only the later has to be true

The following has 2 facts combined (separated by a comma) "My hat on my desk, sucks up dirt.". When you have a sentence made of smaller true parts, how can the sentence be true so long as the smaller parts are true?

We need premises.
Remember the Art Bob Carl example? The conclusion is "Bob murdered Victor.". And it required premises like who said who was in town and knew him etc.

[LOCKSUIT]

"Rocks sleep."

The atoms of this sentence are true. Rocks is true. Sleep is true.

1 1

But the bigger 'part' they build; "Rocks sleep.", is false, there's countless examples ex. "rocks eat" etc.

Following the Course on their site, it suggest the truth of the sentence "Rocks sleep." is:

1

Because both children are 1.

The sum of them is therefore 1.

Right? This is wrong... How does it / does it solve this?

[ivan.moony]

You need predicate logic to express that:

(∀x∊Rocks).Sleep(x)

[LOCKSUIT]

...Huh? How does that figure out if the part they build is true? Premises? You'd need prehand knowledge to determine if it's true.

[ivan.moony]

When you have a whole compound sentence that is true, there are rules that say which parts should be true for the whole sentence to be true. If sentence is composed of only one part [i.e. Sleep(x)] than that part has to be true. Otherwise, the sentence is false.

You may see quantificators as a sentence modifiers that allow variables to range over sets. That is why quantificators itself can't have any truth value assigned - as a parts of a sentence. Only predicates - as a parts of a sentence - can have assigned a truth value.

[LOCKSUIT]

In English, you mean the training data / the premises sets / the base knowledge, is 'marked'/tagged with rank truths, and that to determine one's truth requires modifiers that can validate it and then assign over truth?

[ivan.moony]

We may have a set:

car ∊ Vehicles; train ∊ Vehicles; plane ∊ Vehicles

Now, to say:

(∀x∊Vehicles).Rides(x)

it is the same as we write:

Rides(car) ∧ Rides(train) ∧ Rides(plane)

But if we say:

(∃x∊Vehicles).HasWheels(x)

it is the same as we write:

HasWheels(car) ∨ HasWheels(train) ∨ HasWheels(plane)

[LOCKSUIT]

car, train, plane = vehicle
vehicles = rides, has wheels

New unseen input is: "Car has wheels."

How do we find the sentence is true?

[LOCKSUIT]

On chapter 11

(Need some examples using English sentences, for induction.)
P.S don't use the water boils in case1, case2, case3, hence it always boils....no lol...the below clearly is something deeper going on...

Can someone explain in English an example of the below in the real world? Like Art did showing how we eliminate 2 'nots' in "I'm not going to not pick you.".

I see lots of interesting words like "to match   the antecedent of the premise" "implication introduction" "strip away" and so on, show all this in your reply.



"The relationship holds the other way around as well. Given ∀x.(∃y.p(x,y) ⇒ q(x)), we know that ∀x.∀y.(p(x,y) ⇒ q(x)). We can convert an existential quantifier in the antecedent of an implication into a universal quantifier outside the implication."

"Our proof is shown below. As usual, we start with our premise. We start a subproof by making an assumption. Then we turn the assumption into an existential sentence to match the antecedent of the premise. We use Universal Implication to strip away the quantifier in the premise to expose the implication. Then, we apply Implication Elimination to derive q(x). Finally, we use Implication Introduction, and we generalize using two applications of Universal Introduction."