NAL-5 question

[Robert Johansson]

Hello everyone,

If NARS has learned the following:

<(X * Y) --> (a * a)>.

<(Y * X) --> (a * a)>.

// Derived: <<($1 * $2) --> (a * a)> <=> <($2 * $1) --> (a * a)>>

And also this:

<<(X * Y) --> (a * a)> <=> <(A * B) --> (b * b)>>.

Is there any chance that <(B * A) --> (b * b)> would be derived?

Robert 

[Pei Wang]

Probably not. I don’t see evidence for that conclusion.

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[Tessergon Ng]

TLDR: According to the NAL-4 and NAL-5 rules, it probably would be derived.

In the one hand,  by the premise

<(X * Y) --> (a * a)>.

NARS can get <X --> a>. and <Y --> a>. by compositional rules.

(The definition of products in NAL-4 also supports it)

In the other hand,

we have

<(X * Y) --> (a * a)>.

<<(X * Y) --> (a * a)> <=> <(A * B) --> (b * b)>>.

so NARS can get <(A * B) --> (b * b)>. by NAL-5 detachment.

(Maybe during NAL-1 convertion to get implication <<(X * Y) --> (a * a)> ==> <(A * B) --> (b * b)>> first)

BTW, like the derivation of

<X --> Y>. and <Y --> X>.

NARS can uses premise <(A * B) --> (b * b)> to get conclusion <A --> b> and <B --> b>,

then uses them directly to compose <(B * A) --> (b * b)> directly,

which is not actually related to  <<($1 * $2) --> (a * a)> <=> <($2 * $1) --> (a * a)>>.

(Some detailed steps:

1. begins from common term b, uses <A --> b>. and relation of b-B to compose <(B * A) --> (B * b)> first,

2. and also combine it with relation of b-B to get <(B * b) --> (b * b)>

3. then uses NAL-1 deduction to get <(B * A) --> (b * b)> by reducing common term (B * b)

)

The whole progress just like this.

(The upper task is derived from the downer tasks; Image 2 shows the backward inferencing of NARS)

It can be represented in 2 narsese phrases:

<(*,X,Y) --> (*,a,a)>.

<X --> a>?

<Y --> a>?

<A --> b>.

<B --> b>.

<(*,B,A)-->(*,b,b)>?

And, maybe the processing of NAL in ONA is much simplified, perhaps you can try some alternative versions of NARS such as OpenNARS or PyNARS...

[Pei Wang]

Yes, it makes sense, though the following inference is not currently implemented:

    {<A --> a>, <B --> b>} |-  <(B * A) --> (a * b)>

This is because of control and cost issues, as such a rule can be applied to two arbitrary inheritance statements without a shared term (the shared b in the original example is accidental).

If there is a strong practical need for such a rule, it can be implemented as a special "Structural Rule", using the existence of (B * A) or (a * b) as an additional pre-condition. Anyway, it should be done to two arbitrary inheritance statements, even though it is logically valid.

Regards,

Pei

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