daj95376 wrote:However, I wish he had added one more connector:

(3=9)r3c8 - (9=8)r7c8 - (8=6|7)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2

Yes, no problem, if you write out the relation between all multiple digits.

There are several ways to express the same, all being correct, if it is defined, what means what:

(3=9)r3c8 - (9=8)r7c8 - (8=6|7)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2

(3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (6&7=4|5)r89c2 - (4&5=3)r16c2 (obviously 67r7c1 must mean (6|7)r7c1)

(3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (67=4|5)r89c2 - (45=3)r16c2 (with the convention, that ab in 2 or more cells means a&b)

(3=9)r3c8 - (9=8)r7c8 - (8=67)r7c1 - (67=3)r1689c2 (because of the ALS with 5 digits in 4 cells "NOT both 6&7" or "NOT 3" implies that all other digits must be in the cells - only those of interest are listed)

etc. etc.

daj95376 wrote:There is a single symbol that meets my needs.

"The statement A ∧ B is true if A and B are both true; else it is false".

Why don't you take & then, if you mean it ?

If both <4> and <8> are not true in r13c2, then both <2> and <7> must be true in r13c2.

If both <2> and <7> are not true in r13c2, then both <4> and <8> must be true in r13c2.

This translates into an SL with one special connector being used: ( 4 ∧ 8 = 2 ∧ 7 ) r13c2

You adhere to the same mistake all the time:

"both <4> and <8> are not true in r13c2" means "(NOT 4) AND (NOT 8)", which is equivalent to NOT (4 OR 8) (not the one or the other).

So "If both <4> and <8> are not true in r13c2, then both <2> and <7> must be true in r13c2" has to be written as

(4|8 = 2&7)r13c2

( 4 ∧ 8 = 2 ∧ 7 ) r13c2

would state: "if NOT (4 AND 8) in r13c2 (not both, at least one missing, not the one or not the other), then (2 AND 7)", which is simply wrong.

In other words, it is wrong that the 2 cells must hold either both 4 and 8 or both 2 and 7.

[Added:]What you would need here, is not a new symbol for & between digits, but for the '=' relation, with the following meaning:

E.g. if we take '|=|' for it,

(48 |=| 27)r13c2 means (4|8 = 2&7)r13c2 AND (4&8 = 2|7)r13c2 (note that the second one is different to the first one, read from right to left)

(if you have only 4 candidates in the cells like in the V-loop, both relations are obvious)

If you reach a loop consisting of only such relations, connected with '-' (NAND) relations, the known eliminations of the 2 common digits in all units of the loop cells (and in case of SK loop the digits not present in a pair of the loop cells) are proven (see e.g. blues' way).