Theorem eqimss2 1549

Description: Equality implies the subclass relation.

Assertion

Ref	Expression
eqimss2	⊢ (B = A → A ⊆ B)

Proof of Theorem eqimss2

Step	Hyp	Ref	Expression
1		eqimss 1548	. 2 ⊢ (A = B → A ⊆ B)
2	1	cleqcoms 1104	1 ⊢ (B = A → A ⊆ B)

Syntax hints:   → wi 2   = wceq 1091   ⊆ wss 1487

This theorem is referenced by:  vss 1729  suc11 2341  dmcoeq 2573  oaass 3163  oen0 3165  zorn2 3612

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-in 1491  df-ss 1492

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