Metamath Proof Explorer


Theorem aomclem3

Description: Lemma for dfac11 . Successor case 3, our required well-ordering. (Contributed by Stefan O'Rear, 19-Jan-2015)

Ref Expression
Hypotheses aomclem3.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
aomclem3.c C = a V sup y a R1 dom z B
aomclem3.d D = recs a V C R1 dom z ran a
aomclem3.e E = a b | D -1 a D -1 b
aomclem3.on φ dom z On
aomclem3.su φ dom z = suc dom z
aomclem3.we φ a dom z z a We R1 a
aomclem3.a φ A On
aomclem3.za φ dom z A
aomclem3.y φ a 𝒫 R1 A a y a 𝒫 a Fin
Assertion aomclem3 φ E We R1 dom z

Proof

Step Hyp Ref Expression
1 aomclem3.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 aomclem3.c C = a V sup y a R1 dom z B
3 aomclem3.d D = recs a V C R1 dom z ran a
4 aomclem3.e E = a b | D -1 a D -1 b
5 aomclem3.on φ dom z On
6 aomclem3.su φ dom z = suc dom z
7 aomclem3.we φ a dom z z a We R1 a
8 aomclem3.a φ A On
9 aomclem3.za φ dom z A
10 aomclem3.y φ a 𝒫 R1 A a y a 𝒫 a Fin
11 rneq a = c ran a = ran c
12 11 difeq2d a = c R1 dom z ran a = R1 dom z ran c
13 12 fveq2d a = c C R1 dom z ran a = C R1 dom z ran c
14 13 cbvmptv a V C R1 dom z ran a = c V C R1 dom z ran c
15 recseq a V C R1 dom z ran a = c V C R1 dom z ran c recs a V C R1 dom z ran a = recs c V C R1 dom z ran c
16 14 15 ax-mp recs a V C R1 dom z ran a = recs c V C R1 dom z ran c
17 3 16 eqtri D = recs c V C R1 dom z ran c
18 fvexd φ R1 dom z V
19 1 2 5 6 7 8 9 10 aomclem2 φ a 𝒫 R1 dom z a C a a
20 neeq1 a = d a d
21 fveq2 a = d C a = C d
22 id a = d a = d
23 21 22 eleq12d a = d C a a C d d
24 20 23 imbi12d a = d a C a a d C d d
25 24 cbvralvw a 𝒫 R1 dom z a C a a d 𝒫 R1 dom z d C d d
26 19 25 sylib φ d 𝒫 R1 dom z d C d d
27 17 18 26 4 dnwech φ E We R1 dom z