Metamath Proof Explorer


Theorem aomclem1

Description: Lemma for dfac11 . This is the beginning of the proof that multiple choice is equivalent to choice. Our goal is to construct, by transfinite recursion, a well-ordering of ( R1A ) . In what follows, A is the index of the rank we wish to well-order, z is the collection of well-orderings constructed so far, dom z is the set of ordinal indices of constructed ranks i.e. the next rank to construct, and y is a postulated multiple-choice function.

Successor case 1, define a simple ordering from the well-ordered predecessor. (Contributed by Stefan O'Rear, 18-Jan-2015)

Ref Expression
Hypotheses aomclem1.b B = a b | c R1 dom z c b ¬ c a d R1 dom z d z dom z c d a d b
aomclem1.on φ dom z On
aomclem1.su φ dom z = suc dom z
aomclem1.we φ a dom z z a We R1 a
Assertion aomclem1 φ B Or R1 dom z

Proof

Step Hyp Ref Expression
1 aomclem1.b B = a b | c R1 dom z c b ¬ c a d R1 dom z d z dom z c d a d b
2 aomclem1.on φ dom z On
3 aomclem1.su φ dom z = suc dom z
4 aomclem1.we φ a dom z z a We R1 a
5 fvex R1 dom z V
6 vex z V
7 6 dmex dom z V
8 7 uniex dom z V
9 8 sucid dom z suc dom z
10 9 3 eleqtrrid φ dom z dom z
11 fveq2 a = dom z z a = z dom z
12 fveq2 a = dom z R1 a = R1 dom z
13 11 12 weeq12d a = dom z z a We R1 a z dom z We R1 dom z
14 13 rspcva dom z dom z a dom z z a We R1 a z dom z We R1 dom z
15 10 4 14 syl2anc φ z dom z We R1 dom z
16 1 wepwso R1 dom z V z dom z We R1 dom z B Or 𝒫 R1 dom z
17 5 15 16 sylancr φ B Or 𝒫 R1 dom z
18 3 fveq2d φ R1 dom z = R1 suc dom z
19 onuni dom z On dom z On
20 r1suc dom z On R1 suc dom z = 𝒫 R1 dom z
21 2 19 20 3syl φ R1 suc dom z = 𝒫 R1 dom z
22 18 21 eqtrd φ R1 dom z = 𝒫 R1 dom z
23 soeq2 R1 dom z = 𝒫 R1 dom z B Or R1 dom z B Or 𝒫 R1 dom z
24 22 23 syl φ B Or R1 dom z B Or 𝒫 R1 dom z
25 17 24 mpbird φ B Or R1 dom z