Theorem necompl 3545

Description: A class is not equal to its complement. (Contributed by SF, 11-Jan-2015.)

Assertion

Ref	Expression
necompl	⊢ ∼ A ≠ A

Proof of Theorem necompl

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		pm5.19 349	. . . . 5 ⊢ ¬ (x ∈ A ↔ ¬ x ∈ A)
2		vex 2863	. . . . . . 7 ⊢ x ∈ V
3	2	elcompl 3226	. . . . . 6 ⊢ (x ∈ ∼ A ↔ ¬ x ∈ A)
4	3	bibi2i 304	. . . . 5 ⊢ ((x ∈ A ↔ x ∈ ∼ A) ↔ (x ∈ A ↔ ¬ x ∈ A))
5	1, 4	mtbir 290	. . . 4 ⊢ ¬ (x ∈ A ↔ x ∈ ∼ A)
6		19.8a 1756	. . . 4 ⊢ (¬ (x ∈ A ↔ x ∈ ∼ A) → ∃x ¬ (x ∈ A ↔ x ∈ ∼ A))
7	5, 6	ax-mp 5	. . 3 ⊢ ∃x ¬ (x ∈ A ↔ x ∈ ∼ A)
8		dfcleq 2347	. . . . 5 ⊢ (A = ∼ A ↔ ∀x(x ∈ A ↔ x ∈ ∼ A))
9	8	necon3abii 2547	. . . 4 ⊢ (A ≠ ∼ A ↔ ¬ ∀x(x ∈ A ↔ x ∈ ∼ A))
10		exnal 1574	. . . 4 ⊢ (∃x ¬ (x ∈ A ↔ x ∈ ∼ A) ↔ ¬ ∀x(x ∈ A ↔ x ∈ ∼ A))
11	9, 10	bitr4i 243	. . 3 ⊢ (A ≠ ∼ A ↔ ∃x ¬ (x ∈ A ↔ x ∈ ∼ A))
12	7, 11	mpbir 200	. 2 ⊢ A ≠ ∼ A
13	12	necomi 2599	1 ⊢ ∼ A ≠ A

Syntax hints:  ¬ wn 3   ↔ wb 176  ∀wal 1540  ∃wex 1541   ∈ wcel 1710   ≠ wne 2517   ∼ ccompl 3206

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ne 2519  df-v 2862  df-nin 3212  df-compl 3213

This theorem is referenced by:  nfunv  5139  endisj  6052  ncaddccl  6145  tcdi  6165