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Theorem unss2 4155
Description: Subclass law for union of classes. Exercise 7 of [TakeutiZaring] p. 18. (Contributed by NM, 14-Oct-1999.)
Assertion
Ref Expression
unss2 (𝐴𝐵 → (𝐶𝐴) ⊆ (𝐶𝐵))

Proof of Theorem unss2
StepHypRef Expression
1 unss1 4153 . 2 (𝐴𝐵 → (𝐴𝐶) ⊆ (𝐵𝐶))
2 uncom 4127 . 2 (𝐶𝐴) = (𝐴𝐶)
3 uncom 4127 . 2 (𝐶𝐵) = (𝐵𝐶)
41, 2, 33sstr4g 4010 1 (𝐴𝐵 → (𝐶𝐴) ⊆ (𝐶𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  cun 3932  wss 3934
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1905  ax-6 1964  ax-7 2009  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2154  ax-12 2170  ax-ext 2791
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1534  df-ex 1775  df-nf 1779  df-sb 2064  df-clab 2798  df-cleq 2812  df-clel 2891  df-nfc 2961  df-v 3495  df-un 3939  df-in 3941  df-ss 3950
This theorem is referenced by:  unss12  4156  ord3ex  5278  xpider  8360  fin1a2lem13  9826  canthp1lem2  10067  seqexw  13377  uniioombllem3  24178  volcn  24199  dvres2lem  24500  bnj1413  32300  bnj1408  32301
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