Metamath Proof Explorer


Theorem 1sdom

Description: A set that strictly dominates ordinal 1 has at least 2 different members. (Closely related to 2dom .) (Contributed by Mario Carneiro, 12-Jan-2013)

Ref Expression
Assertion 1sdom A V 1 𝑜 A x A y A ¬ x = y

Proof

Step Hyp Ref Expression
1 breq2 a = A 1 𝑜 a 1 𝑜 A
2 rexeq a = A y a ¬ x = y y A ¬ x = y
3 2 rexeqbi1dv a = A x a y a ¬ x = y x A y A ¬ x = y
4 1onn 1 𝑜 ω
5 sucdom 1 𝑜 ω 1 𝑜 a suc 1 𝑜 a
6 4 5 ax-mp 1 𝑜 a suc 1 𝑜 a
7 df-2o 2 𝑜 = suc 1 𝑜
8 7 breq1i 2 𝑜 a suc 1 𝑜 a
9 2dom 2 𝑜 a x a y a ¬ x = y
10 df2o3 2 𝑜 = 1 𝑜
11 vex x V
12 vex y V
13 0ex V
14 1oex 1 𝑜 V
15 11 12 13 14 funpr x y Fun x y 1 𝑜
16 df-ne x y ¬ x = y
17 1n0 1 𝑜
18 17 necomi 1 𝑜
19 13 14 11 12 fpr 1 𝑜 x 1 𝑜 y : 1 𝑜 x y
20 18 19 ax-mp x 1 𝑜 y : 1 𝑜 x y
21 df-f1 x 1 𝑜 y : 1 𝑜 1-1 x y x 1 𝑜 y : 1 𝑜 x y Fun x 1 𝑜 y -1
22 20 21 mpbiran x 1 𝑜 y : 1 𝑜 1-1 x y Fun x 1 𝑜 y -1
23 13 11 cnvsn x -1 = x
24 14 12 cnvsn 1 𝑜 y -1 = y 1 𝑜
25 23 24 uneq12i x -1 1 𝑜 y -1 = x y 1 𝑜
26 df-pr x 1 𝑜 y = x 1 𝑜 y
27 26 cnveqi x 1 𝑜 y -1 = x 1 𝑜 y -1
28 cnvun x 1 𝑜 y -1 = x -1 1 𝑜 y -1
29 27 28 eqtri x 1 𝑜 y -1 = x -1 1 𝑜 y -1
30 df-pr x y 1 𝑜 = x y 1 𝑜
31 25 29 30 3eqtr4i x 1 𝑜 y -1 = x y 1 𝑜
32 31 funeqi Fun x 1 𝑜 y -1 Fun x y 1 𝑜
33 22 32 bitr2i Fun x y 1 𝑜 x 1 𝑜 y : 1 𝑜 1-1 x y
34 15 16 33 3imtr3i ¬ x = y x 1 𝑜 y : 1 𝑜 1-1 x y
35 prssi x a y a x y a
36 f1ss x 1 𝑜 y : 1 𝑜 1-1 x y x y a x 1 𝑜 y : 1 𝑜 1-1 a
37 34 35 36 syl2an ¬ x = y x a y a x 1 𝑜 y : 1 𝑜 1-1 a
38 prex x 1 𝑜 y V
39 f1eq1 f = x 1 𝑜 y f : 1 𝑜 1-1 a x 1 𝑜 y : 1 𝑜 1-1 a
40 38 39 spcev x 1 𝑜 y : 1 𝑜 1-1 a f f : 1 𝑜 1-1 a
41 37 40 syl ¬ x = y x a y a f f : 1 𝑜 1-1 a
42 vex a V
43 42 brdom 1 𝑜 a f f : 1 𝑜 1-1 a
44 41 43 sylibr ¬ x = y x a y a 1 𝑜 a
45 44 expcom x a y a ¬ x = y 1 𝑜 a
46 45 rexlimivv x a y a ¬ x = y 1 𝑜 a
47 10 46 eqbrtrid x a y a ¬ x = y 2 𝑜 a
48 9 47 impbii 2 𝑜 a x a y a ¬ x = y
49 6 8 48 3bitr2i 1 𝑜 a x a y a ¬ x = y
50 1 3 49 vtoclbg A V 1 𝑜 A x A y A ¬ x = y