Theorem ss2rab 1553

Description: Equivalence of restricted abstraction subclass and implication.

Assertion

Ref	Expression
ss2rab	|- ({x e. A | ph} (_ {x e. A | ps} <-> A.x e. A (ph -> ps))

Proof of Theorem ss2rab

Step	Hyp	Ref	Expression
1		df-rab 1208	. . 3 |- {x e. A | ph} = {x | (x e. A /\ ph)}
2		df-rab 1208	. . 3 |- {x e. A | ps} = {x | (x e. A /\ ps)}
3	1, 2	sseq12i 1526	. 2 |- ({x e. A | ph} (_ {x e. A | ps} <-> {x | (x e. A /\ ph)} (_ {x | (x e. A /\ ps)})
4		ss2ab 1551	. 2 |- ({x | (x e. A /\ ph)} (_ {x | (x e. A /\ ps)} <-> A.x((x e. A /\ ph) -> (x e. A /\ ps)))
5		df-ral 1205	. . 3 |- (A.x e. A (ph -> ps) <-> A.x(x e. A -> (ph -> ps)))
6		imdistan 339	. . . 4 |- ((x e. A -> (ph -> ps)) <-> ((x e. A /\ ph) -> (x e. A /\ ps)))
7	6	bial 695	. . 3 |- (A.x(x e. A -> (ph -> ps)) <-> A.x((x e. A /\ ph) -> (x e. A /\ ps)))
8	5, 7	bitr2 152	. 2 |- (A.x((x e. A /\ ph) -> (x e. A /\ ps)) <-> A.x e. A (ph -> ps))
9	3, 4, 8	3bitr 155	1 |- ({x e. A | ph} (_ {x e. A | ps} <-> A.x e. A (ph -> ps))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196  A.wal 672  {cab 1090   e. wcel 1092  A.wral 1201  {crab 1204   (_ wss 1487

This theorem is referenced by:  ss2rabi 1554  rankr1id 3539  scottex 3541  ondomon 3662  uzwo3lem1 4614  uzwo3lem2 4615  occont 5168  hsupss 5310  spanss 5319  chpssat 5756

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-ral 1205  df-rab 1208  df-in 1491  df-ss 1492

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