Theorem sseqin2 1656

Description: A relationship between subclass and intersection. Similar to Exercise 9 of [TakeutiZaring] p. 18.

Assertion

Ref	Expression
sseqin2	⊢ (A ⊆ B ↔ (B ∩ A) = A)

Proof of Theorem sseqin2

Step	Hyp	Ref	Expression
1		df-ss 1492	. 2 ⊢ (A ⊆ B ↔ (A ∩ B) = A)
2		incom 1636	. . 3 ⊢ (A ∩ B) = (B ∩ A)
3	2	cleq1i 1108	. 2 ⊢ ((A ∩ B) = A ↔ (B ∩ A) = A)
4	1, 3	bitr 151	1 ⊢ (A ⊆ B ↔ (B ∩ A) = A)

Syntax hints:   ↔ wb 127   = wceq 1091   ∩ cin 1486   ⊆ wss 1487

This theorem is referenced by:  dfss4 1667  onfr 2237  resabs1 2592  pw2en 3348  fiint 3445  kmlem10 3589  cmcmlem 5500  ssmd2 5735

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  df-ss 1492

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