Theorem eirr 3450

Description: No class is a member of itself. Exercise 6 of [TakeutiZaring] p. 22.

Assertion

Ref	Expression
eirr	⊢ ¬ A ∈ A

Proof of Theorem eirr

Step	Hyp	Ref	Expression
1		eleq2 1150	. . . . 5 ⊢ (x = A → (x ∈ x ↔ x ∈ A))
2		eleq1 1149	. . . . 5 ⊢ (x = A → (x ∈ A ↔ A ∈ A))
3	1, 2	bitrd 406	. . . 4 ⊢ (x = A → (x ∈ x ↔ A ∈ A))
4	3	negbid 463	. . 3 ⊢ (x = A → (¬ x ∈ x ↔ ¬ A ∈ A))
5		eirrv 3449	. . 3 ⊢ ¬ x ∈ x
6	4, 5	vtoclg 1383	. 2 ⊢ (A ∈ A → ¬ A ∈ A)
7		pm2.01 80	. 2 ⊢ ((A ∈ A → ¬ A ∈ A) → ¬ A ∈ A)
8	6, 7	ax-mp 6	1 ⊢ ¬ A ∈ A

Syntax hints:  ¬ wn 1   → wi 2   ∈ wel 803   = wceq 1091   ∈ wcel 1092

This theorem is referenced by:  sucprcreg 3451

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077  ax-reg 1078

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812

metamath.org