Lemma 7 - Revision history

Tomtom2357 at 21:40, 3 December 2016

2016-12-03T21:40:40Z

← Older revision		Revision as of 14:40, 3 December 2016
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	==Lemma 7:==		==Lemma 7:==
	This proof is much longer than anticipated (and it's still not completed), ~~but~~ it will be split into six cases (with each case relying on the previous cases):		This proof is much longer than anticipated (and it's still not completed), so it will be split into six cases (with each case relying on the previous cases):

	[[Lemma 7.1]]: If there are two size 5 sets intersecting at 4 elements, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.1]]: If there are two size 5 sets intersecting at 4 elements, then <math>\mathcal{A}</math> is Frankl's

Tomtom2357: /* Lemma 7: */

2016-12-03T01:55:32Z

Lemma 7:

← Older revision		Revision as of 18:55, 2 December 2016
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	[[Lemma 7.3]]: If there are two size 5 sets intersecting at 2 elements, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.3]]: If there are two size 5 sets intersecting at 2 elements, then <math>\mathcal{A}</math> is Frankl's

	[[Lemma 7.4]]: If there are two size 5 sets ~~intersecting in 1 element~~, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.4]]: If there are two intersecting size 5 sets, then <math>\mathcal{A}</math> is Frankl's

	[[Lemma 7.5]]: If there are two size 5 sets, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.5]]: If there are two size 5 sets, then <math>\mathcal{A}</math> is Frankl's

	[[Lemma 7.6]]: If there is a size 5 set, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.6]]: If there is a size 5 set, then <math>\mathcal{A}</math> is Frankl's

Tomtom2357: /* Lemma 7: */

2016-12-02T00:12:55Z

Lemma 7:

← Older revision		Revision as of 17:12, 1 December 2016
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	[[Lemma 7.1]]: If there are two size 5 sets intersecting at 4 elements, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.1]]: If there are two size 5 sets intersecting at 4 elements, then <math>\mathcal{A}</math> is Frankl's

	[[Lemma 7.1]]: If there are two size 5 sets intersecting at 3 elements, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.2]]: If there are two size 5 sets intersecting at 3 elements, then <math>\mathcal{A}</math> is Frankl's

	[[Lemma 7.1]]: If there are two size 5 sets intersecting at 2 elements, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.3]]: If there are two size 5 sets intersecting at 2 elements, then <math>\mathcal{A}</math> is Frankl's

	[[Lemma 7.1]]: If there are two size 5 sets intersecting in 1 element, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.4]]: If there are two size 5 sets intersecting in 1 element, then <math>\mathcal{A}</math> is Frankl's

	[[Lemma 7.1]]: If there are two size 5 sets, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.5]]: If there are two size 5 sets, then <math>\mathcal{A}</math> is Frankl's

	[[Lemma 7.1]]: If there is a size 5 set, then <math>\mathcal{A}</math> is Frankl's		[[Lemma 7.6]]: If there is a size 5 set, then <math>\mathcal{A}</math> is Frankl's

Tomtom2357: Created page, split into sections

2016-12-01T03:36:25Z

Created page, split into sections

New page

This page proves a lemma for the [[m=13 case of FUNC]].

==Lemma 7:==
This proof is much longer than anticipated (and it's still not completed), but it will be split into six cases (with each case relying on the previous cases):

[[Lemma 7.1]]: If there are two size 5 sets intersecting at 4 elements, then <math>\mathcal{A}</math> is Frankl's

[[Lemma 7.1]]: If there are two size 5 sets intersecting at 3 elements, then <math>\mathcal{A}</math> is Frankl's

[[Lemma 7.1]]: If there are two size 5 sets intersecting at 2 elements, then <math>\mathcal{A}</math> is Frankl's

[[Lemma 7.1]]: If there are two size 5 sets intersecting in 1 element, then <math>\mathcal{A}</math> is Frankl's

[[Lemma 7.1]]: If there are two size 5 sets, then <math>\mathcal{A}</math> is Frankl's

[[Lemma 7.1]]: If there is a size 5 set, then <math>\mathcal{A}</math> is Frankl's