Theorem difab 1693

Description: Difference of two class abstractions.

Assertion

Ref	Expression
difab	|- ({x | ph} \ {x | ps}) = {x | (ph /\ -. ps)}

Proof of Theorem difab

Step	Hyp	Ref	Expression
1		sbn 882	. . . . . 6 |- ([y / x] -. ps <-> -. [y / x]ps)
2		df-clab 1093	. . . . . 6 |- (y e. {x | -. ps} <-> [y / x] -. ps)
3		df-clab 1093	. . . . . . 7 |- (y e. {x | ps} <-> [y / x]ps)
4	3	negbii 162	. . . . . 6 |- (-. y e. {x | ps} <-> -. [y / x]ps)
5	1, 2, 4	3bitr4 158	. . . . 5 |- (y e. {x | -. ps} <-> -. y e. {x | ps})
6		visset 1350	. . . . . 6 |- y e. V
7	6	biantrur 544	. . . . 5 |- (-. y e. {x | ps} <-> (y e. V /\ -. y e. {x | ps}))
8	5, 7	bitr2 152	. . . 4 |- ((y e. V /\ -. y e. {x | ps}) <-> y e. {x | -. ps})
9	8	difeqri 1589	. . 3 |- (V \ {x | ps}) = {x | -. ps}
10	9	ineq2i 1642	. 2 |- ({x | ph} i^i (V \ {x | ps})) = ({x | ph} i^i {x | -. ps})
11		invdif 1674	. 2 |- ({x | ph} i^i (V \ {x | ps})) = ({x | ph} \ {x | ps})
12		inab 1692	. 2 |- ({x | ph} i^i {x | -. ps}) = {x | (ph /\ -. ps)}
13	10, 11, 12	3eqtr3 1124	1 |- ({x | ph} \ {x | ps}) = {x | (ph /\ -. ps)}

Syntax hints:  -. wn 1   /\ wa 196  [wsb 852  {cab 1090   = wceq 1091   e. wcel 1092  Vcvv 1348   \ cdif 1484   i^i 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

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