Theorem difab 1693

Description: Difference of two class abstractions.

Assertion

Ref	Expression
difab	⊢ ({x∣φ} ∖ {x∣ψ}) = {x∣(φ ∧ ¬ ψ)}

Proof of Theorem difab

Step	Hyp	Ref	Expression
1		sbn 882	. . . . . 6 ⊢ ([y / x] ¬ ψ ↔ ¬ [y / x]ψ)
2		df-clab 1093	. . . . . 6 ⊢ (y ∈ {x∣ ¬ ψ} ↔ [y / x] ¬ ψ)
3		df-clab 1093	. . . . . . 7 ⊢ (y ∈ {x∣ψ} ↔ [y / x]ψ)
4	3	negbii 162	. . . . . 6 ⊢ (¬ y ∈ {x∣ψ} ↔ ¬ [y / x]ψ)
5	1, 2, 4	3bitr4 158	. . . . 5 ⊢ (y ∈ {x∣ ¬ ψ} ↔ ¬ y ∈ {x∣ψ})
6		visset 1350	. . . . . 6 ⊢ y ∈ V
7	6	biantrur 544	. . . . 5 ⊢ (¬ y ∈ {x∣ψ} ↔ (y ∈ V ∧ ¬ y ∈ {x∣ψ}))
8	5, 7	bitr2 152	. . . 4 ⊢ ((y ∈ V ∧ ¬ y ∈ {x∣ψ}) ↔ y ∈ {x∣ ¬ ψ})
9	8	difeqri 1589	. . 3 ⊢ (V ∖ {x∣ψ}) = {x∣ ¬ ψ}
10	9	ineq2i 1642	. 2 ⊢ ({x∣φ} ∩ (V ∖ {x∣ψ})) = ({x∣φ} ∩ {x∣ ¬ ψ})
11		invdif 1674	. 2 ⊢ ({x∣φ} ∩ (V ∖ {x∣ψ})) = ({x∣φ} ∖ {x∣ψ})
12		inab 1692	. 2 ⊢ ({x∣φ} ∩ {x∣ ¬ ψ}) = {x∣(φ ∧ ¬ ψ)}
13	10, 11, 12	3eqtr3 1124	1 ⊢ ({x∣φ} ∖ {x∣ψ}) = {x∣(φ ∧ ¬ ψ)}

Syntax hints:  ¬ wn 1   ∧ wa 196  [wsb 852  {cab 1090   = wceq 1091   ∈ wcel 1092  Vcvv 1348   ∖ cdif 1484   ∩ cin 1486

This theorem is referenced by:  difrab 1695

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-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-dif 1489  df-in 1491

metamath.org