Theorem difab 2269

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 1231	. . . . . 6 |- ([y / x] -. ps <-> -. [y / x]ps)
2		df-clab 1464	. . . . . 6 |- (y e. {x | -. ps} <-> [y / x] -. ps)
3		df-clab 1464	. . . . . . 7 |- (y e. {x | ps} <-> [y / x]ps)
4	3	negbii 187	. . . . . 6 |- (-. y e. {x | ps} <-> -. [y / x]ps)
5	1, 2, 4	3bitr4 183	. . . . 5 |- (y e. {x | -. ps} <-> -. y e. {x | ps})
6		visset 1813	. . . . . 6 |- y e. V
7	6	biantrur 725	. . . . 5 |- (-. y e. {x | ps} <-> (y e. V /\ -. y e. {x | ps}))
8	5, 7	bitr2 174	. . . 4 |- ((y e. V /\ -. y e. {x | ps}) <-> y e. {x | -. ps})
9	8	difeqri 2160	. . 3 |- (V \ {x | ps}) = {x | -. ps}
10	9	ineq2i 2214	. 2 |- ({x | ph} i^i (V \ {x | ps})) = ({x | ph} i^i {x | -. ps})
11		invdif 2249	. 2 |- ({x | ph} i^i (V \ {x | ps})) = ({x | ph} \ {x | ps})
12		inab 2268	. 2 |- ({x | ph} i^i {x | -. ps}) = {x | (ph /\ -. ps)}
13	10, 11, 12	3eqtr3 1503	1 |- ({x | ph} \ {x | ps}) = {x | (ph /\ -. ps)}

Syntax hints:  -. wn 2   /\ wa 223   = wceq 956   e. wcel 958  [wsbc 1170  {cab 1463  Vcvv 1811   \ cdif 2044   i^i cin 2046

This theorem is referenced by:  difrab 2273

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-v 1812  df-dif 2049  df-in 2051

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