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Triggered by previous question, can one prove GCD(0,8)≠1 purely by lattice laws?

Brute force Prover9/Mace4 assertions

x ^ y = y ^ x. 
(x ^ y) ^ z = x ^ (y ^ z). 
x ^ (x v y) = x. 
x v y = y v x. 
(x v y) v z = x v (y v z). 
x v (x ^ y) = x.

1 v x = x.
1 ^ x = 1.
0 ^ x = 1.

exhibit no [finite] model, which is indication that the system is inconsistent. I have trouble, however, understanding how to elevate this intuition into a formal proof (there is no goal).

Community
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Tegiri Nenashi
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  • Do you have x ^ x = x from your laws ? – mercio Mar 18 '11 at 21:20
  • @chandok: Yes, it is redundant. – Tegiri Nenashi Mar 18 '11 at 21:28
  • Is 0 v x = 1 correct? – Myself Mar 19 '11 at 01:24
  • That is what the original thread suggested. It has been refuted together with another option: 0 v x = 1. The correct version is 0 v x = x and, predictably, Mace4 generates 2 element model. I'm interested, however disproving the wrong assertion 0 v x = 1. I tried putting its negation into Prover9 goal, still Prover9 doesn't seem to be able to derive it. – Tegiri Nenashi Mar 19 '11 at 01:36
  • @Tegiri: You either have things backward, or you are viewing the divisibility order backwards, with $a|b$ meaning $b\leq a$... – Arturo Magidin Mar 19 '11 at 02:00
  • @Arturo: distinguishing left and right is extraordinary ability:-) I agree that having GCD lattice order being compatible with standard total order is a natural choice. – Tegiri Nenashi Mar 19 '11 at 17:47
  • @Arturo: Edited: swapped join and meet to eliminate confusion. – Tegiri Nenashi Mar 19 '11 at 17:53
  • @Tegiri: Your listed rules are not inconsistent: they havee a model with one element, all binary operations the obvious ones, and all nullary operations the unique element. – Arturo Magidin Mar 19 '11 at 23:03

2 Answers2

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Note $\rm\ x = 0\ $ in $\rm\ x \wedge (x \vee y)\ =\ x\ \ \Rightarrow\ \ 0\wedge (0\vee y)\ =\ 0\ \ $ contra $\rm\ \ 0\wedge x\ =\ 1\ \ $ (presuming $\rm\ 0 \ne 1\:$).

Alternatively, recall that the idempotent laws follows from the absorption laws, viz.

$$\rm x\wedge x\ =\ x\wedge (x\vee (x\wedge x))\ =\ x $$

Hence $\rm\ \ 0\wedge 0\ =\ 0\ \ $ contra $\rm\ \ 0\wedge x\ =\ 1\:.$

Bill Dubuque
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  • Just to clarify: you are using $0$ and $1$ to denote the corresponding integers, and not to denote the infimum and supremum of the lattice, correct? – Arturo Magidin Mar 19 '11 at 21:21
  • No, the above is in the equational theory that the OP gave. – Bill Dubuque Mar 19 '11 at 21:37
  • @Bill: Right: so that $0$ denotes the usual $0$, not the least element of the lattice. (His theory includes, e.g., $1\lor x = x$, which of course means that "1" does not represent the maximum, as is standard in 01-lattices). – Arturo Magidin Mar 19 '11 at 21:39
  • @Arturo: Possible models are not relevant to the above equational proof (other than the assumption that $\rm\ 0 \ne 1:$). – Bill Dubuque Mar 19 '11 at 21:57
  • I tried absorption and even distributivity, but this rather simple proof escaped me. Thank you. – Tegiri Nenashi Mar 21 '11 at 19:21
  • Here is automatic proof, which is identical to Bill's. The culprit was to go after correct goal! 1 0 ^ x = 0 # label(non_clause) # label(goal). [goal]. 4 x ^ (x v y) = x. [assumption]. 10 0 ^ x = 1. [assumption]. 11 0 ^ c1 != 0. [deny(1)]. 12 0 != 1. [copy(11),rewrite([10(3)]),flip(a)]. 30 0 = 1. [para(10(a,1),4(a,1)),flip(a)]. 31 $F. [resolve(30,a,12,a)]. – Tegiri Nenashi Mar 21 '11 at 19:39
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Solved the remaining bit of the puzzle: GCD(0,8)≠0.

1 v x = x ⇒  1 v 0 = 0
0 ^ x = 0 ⇒  1 ^ 0 = 0  
1 v 0 = 1 ^ 0 ⇒  1 = 0
Tegiri Nenashi
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