Theorem difab 3524

Description: Difference of two class abstractions. (Contributed by NM, 23-Oct-2004.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

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

Proof of Theorem difab

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-clab 2340	. . 3 |- (y e. {x | (ph /\ -. ps)} <-> [y / x](ph /\ -. ps))
2		sban 2069	. . 3 |- ([y / x](ph /\ -. ps) <-> ([y / x]ph /\ [y / x] -. ps))
3		df-clab 2340	. . . . 5 |- (y e. {x | ph} <-> [y / x]ph)
4	3	bicomi 193	. . . 4 |- ([y / x]ph <-> y e. {x | ph})
5		sbn 2062	. . . . 5 |- ([y / x] -. ps <-> -. [y / x]ps)
6		df-clab 2340	. . . . 5 |- (y e. {x | ps} <-> [y / x]ps)
7	5, 6	xchbinxr 302	. . . 4 |- ([y / x] -. ps <-> -. y e. {x | ps})
8	4, 7	anbi12i 678	. . 3 |- (([y / x]ph /\ [y / x] -. ps) <-> (y e. {x | ph} /\ -. y e. {x | ps}))
9	1, 2, 8	3bitrri 263	. 2 |- ((y e. {x | ph} /\ -. y e. {x | ps}) <-> y e. {x | (ph /\ -. ps)})
10	9	difeqri 3388	1 |- ({x | ph} \ {x | ps}) = {x | (ph /\ -. ps)}

Syntax hints:  -. wn 3   /\ wa 358   = wceq 1642  [wsb 1648   e. wcel 1710  {cab 2339   \ cdif 3207

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-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216

This theorem is referenced by:  notab  3526  difrab  3530  notrab  3533