Metamath Proof Explorer


Theorem minregex2

Description: Given any cardinal number A , there exists an argument x , which yields the least regular uncountable value of aleph which dominates A . This proof uses AC. (Contributed by RP, 24-Nov-2023)

Ref Expression
Assertion minregex2 A ran card ω x On x = y On | y A y cf y = y

Proof

Step Hyp Ref Expression
1 minregex A ran card ω x On x = y On | y A y cf y = y
2 eldifi A ran card ω A ran card
3 iscard4 card A = A A ran card
4 2 3 sylibr A ran card ω card A = A
5 4 adantr A ran card ω y On card A = A
6 alephcard card y = y
7 6 a1i A ran card ω y On card y = y
8 5 7 sseq12d A ran card ω y On card A card y A y
9 numth3 A ran card ω A dom card
10 alephon y On
11 onenon y On y dom card
12 10 11 mp1i y On y dom card
13 carddom2 A dom card y dom card card A card y A y
14 9 12 13 syl2an A ran card ω y On card A card y A y
15 8 14 bitr3d A ran card ω y On A y A y
16 15 3anbi2d A ran card ω y On y A y cf y = y y A y cf y = y
17 16 rabbidva A ran card ω y On | y A y cf y = y = y On | y A y cf y = y
18 17 inteqd A ran card ω y On | y A y cf y = y = y On | y A y cf y = y
19 18 eqeq2d A ran card ω x = y On | y A y cf y = y x = y On | y A y cf y = y
20 19 rexbidv A ran card ω x On x = y On | y A y cf y = y x On x = y On | y A y cf y = y
21 1 20 mpbid A ran card ω x On x = y On | y A y cf y = y