Theorem inab 1692

Description: Intersection of two class abstractions.

Assertion

Ref	Expression
inab	|- ({x | ph} i^i {x | ps}) = {x | (ph /\ ps)}

Proof of Theorem inab

Step	Hyp	Ref	Expression
1		df-clab 1093	. . . . 5 |- (y e. {x | ph} <-> [y / x]ph)
2		df-clab 1093	. . . . 5 |- (y e. {x | ps} <-> [y / x]ps)
3	1, 2	anbi12i 369	. . . 4 |- ((y e. {x | ph} /\ y e. {x | ps}) <-> ([y / x]ph /\ [y / x]ps))
4		sban 889	. . . 4 |- ([y / x](ph /\ ps) <-> ([y / x]ph /\ [y / x]ps))
5	3, 4	bitr4 154	. . 3 |- ((y e. {x | ph} /\ y e. {x | ps}) <-> [y / x](ph /\ ps))
6		elin 1635	. . 3 |- (y e. ({x | ph} i^i {x | ps}) <-> (y e. {x | ph} /\ y e. {x | ps}))
7		df-clab 1093	. . 3 |- (y e. {x | (ph /\ ps)} <-> [y / x](ph /\ ps))
8	5, 6, 7	3bitr4 158	. 2 |- (y e. ({x | ph} i^i {x | ps}) <-> y e. {x | (ph /\ ps)})
9	8	cleqri 1101	1 |- ({x | ph} i^i {x | ps}) = {x | (ph /\ ps)}

Syntax hints:   /\ wa 196  [wsb 852  {cab 1090   = wceq 1091   e. wcel 1092   i^i cin 1486

This theorem is referenced by:  difab 1693  dfrab2 1696  ssenen 3399

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-in 1491

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