Can someone a little more fluent in boolean algebra post the transformation that gets you from (1-8) to (1-9) (pg 107-108 in the pdf)? I haven’t been able to work it out.
A different question about 1-8. I was able to figure out how he got A!B = !B (where ! is bar) but using the Boolean identities he provides, I couldn’t get to B!A = !A. Can anyone enlighten me on this?
It seems we have just one rule to eliminate variables: substitution. For example, given A=BC and BCD=E, we can eliminate BC by substituting A for BC in BCD=E. Thus, we must have equation !A=X to get to B!A=!A, and to get to !A=X we must have !A=Y, and so on.
So it seems impossible in given axiomatic system to derive B!A=!A from !B=AD. Am I missing something?
EDIT: Here I take axioms in 1.12 as a basis for proposition calculus and I don’t use any interpretation of them.
I think you just have to prove by truth table that P = P + PQ for any P, Q. Then we have:
A = A + AD = A + !B ⇒ !A = !A !!B = !A B
Second equation is by truth table, third is by definition. The equation after the implication is by DeMorgan’s law (Jaynes calls it “duality”) and the last is by double negation.
True+X=True and (True X)=X do follow from textual description of conjuction and dusjunction, but text before 1.13 suggests (misleads?) that B!A=!A can be derived by using axioms 1.12 only. Latter seems impossible.
Apologies, I totally edited out that part of my comment after finding a much simpler proof. I think my new derivation is fine assuming that proof by truth table is valid (which should be uncontroversial given that he uses it soon after this to show that AND and NOT are an “adequate set” for representing every logic function).
Edit: I was not thinking clearly above. Of course proof by truth table is valid, because truth tables are the basis of the notion of logical equality, and Jayne’s axioms don’t make sense unless you accept that notion as a given.
Actually, you were thinking clearly. We can interpret 1.12 as axioms of proposition calculus, in a strange form of course. As I’ve done partly because of not very rigorous narration.
I’m not sure there is a transformation intended… Proposition (1-9) appears by itself in a section that discusses “implication” (it introduces ⇒ as a shorthand for A=AB) and does not appear to follow from (1-8).
Can someone a little more fluent in boolean algebra post the transformation that gets you from (1-8) to (1-9) (pg 107-108 in the pdf)? I haven’t been able to work it out.
A different question about 1-8. I was able to figure out how he got A!B = !B (where ! is bar) but using the Boolean identities he provides, I couldn’t get to B!A = !A. Can anyone enlighten me on this?
It seems we have just one rule to eliminate variables: substitution. For example, given A=BC and BCD=E, we can eliminate BC by substituting A for BC in BCD=E. Thus, we must have equation !A=X to get to B!A=!A, and to get to !A=X we must have !A=Y, and so on.
So it seems impossible in given axiomatic system to derive B!A=!A from !B=AD. Am I missing something?
EDIT: Here I take axioms in 1.12 as a basis for proposition calculus and I don’t use any interpretation of them.
Perhaps what is missing is these rules:
AT = A (1)
AF = F (2)
A + T = T (3)
A + F = A (4)
Which can be derived from the given axioms, apparently. I’m not sure if some necessary axioms were omitted.
Using some of these, here’s one way to derive B!A=!A from !B=AD:
!B = AD
!B + A = AD + A
!B + A = AD + AT (1)
!B + A = A(D + T) (Distributivity)
!B + A = AT (3)
!B + A = A (1)
!!B!A = !A (Duality)
B!A = !A
I think you just have to prove by truth table that P = P + PQ for any P, Q. Then we have:
A = A + AD = A + !B ⇒ !A = !A !!B = !A B
Second equation is by truth table, third is by definition. The equation after the implication is by DeMorgan’s law (Jaynes calls it “duality”) and the last is by double negation.
True+X=True and (True X)=X do follow from textual description of conjuction and dusjunction, but text before 1.13 suggests (misleads?) that B!A=!A can be derived by using axioms 1.12 only. Latter seems impossible.
Apologies, I totally edited out that part of my comment after finding a much simpler proof. I think my new derivation is fine assuming that proof by truth table is valid (which should be uncontroversial given that he uses it soon after this to show that AND and NOT are an “adequate set” for representing every logic function).
Edit: I was not thinking clearly above. Of course proof by truth table is valid, because truth tables are the basis of the notion of logical equality, and Jayne’s axioms don’t make sense unless you accept that notion as a given.
Actually, you were thinking clearly. We can interpret 1.12 as axioms of proposition calculus, in a strange form of course. As I’ve done partly because of not very rigorous narration.
I think that’s a truly excellent question, BTW. (If someone has an authoritative answer it should probably be ROT13′d.)
PS—in the free pdf it’s 1-8. In the book the problem seems to have been renumbered to 1.13
I’m not sure there is a transformation intended… Proposition (1-9) appears by itself in a section that discusses “implication” (it introduces ⇒ as a shorthand for A=AB) and does not appear to follow from (1-8).
Hrm, no wonder it didn’t work out. Thanks.