Metamath Proof Explorer


Theorem aomclem7

Description: Lemma for dfac11 . ( R1A ) is well-orderable. (Contributed by Stefan O'Rear, 20-Jan-2015)

Ref Expression
Hypotheses aomclem6.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
aomclem6.c C = a V sup y a R1 dom z B
aomclem6.d D = recs a V C R1 dom z ran a
aomclem6.e E = a b | D -1 a D -1 b
aomclem6.f F = a b | rank a E rank b rank a = rank b a z suc rank a b
aomclem6.g G = if dom z = dom z F E R1 dom z × R1 dom z
aomclem6.h H = recs z V G
aomclem6.a φ A On
aomclem6.y φ a 𝒫 R1 A a y a 𝒫 a Fin
Assertion aomclem7 φ b b We R1 A

Proof

Step Hyp Ref Expression
1 aomclem6.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 aomclem6.c C = a V sup y a R1 dom z B
3 aomclem6.d D = recs a V C R1 dom z ran a
4 aomclem6.e E = a b | D -1 a D -1 b
5 aomclem6.f F = a b | rank a E rank b rank a = rank b a z suc rank a b
6 aomclem6.g G = if dom z = dom z F E R1 dom z × R1 dom z
7 aomclem6.h H = recs z V G
8 aomclem6.a φ A On
9 aomclem6.y φ a 𝒫 R1 A a y a 𝒫 a Fin
10 1 2 3 4 5 6 7 8 9 aomclem6 φ H A We R1 A
11 fvex H A V
12 weeq1 b = H A b We R1 A H A We R1 A
13 11 12 spcev H A We R1 A b b We R1 A
14 10 13 syl φ b b We R1 A