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Theorem csbrn 6106
Description: Distribute proper substitution through the range of a class. (Contributed by Alan Sare, 10-Nov-2012.)
Assertion
Ref Expression
csbrn 𝐴 / 𝑥ran 𝐵 = ran 𝐴 / 𝑥𝐵

Proof of Theorem csbrn
StepHypRef Expression
1 csbima12 5987 . . 3 𝐴 / 𝑥(𝐵 “ V) = (𝐴 / 𝑥𝐵𝐴 / 𝑥V)
2 csbconstg 3851 . . . . 5 (𝐴 ∈ V → 𝐴 / 𝑥V = V)
32imaeq2d 5969 . . . 4 (𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V))
4 0ima 5986 . . . . . 6 (∅ “ V) = ∅
54eqcomi 2747 . . . . 5 ∅ = (∅ “ V)
6 csbprc 4340 . . . . . . 7 𝐴 ∈ V → 𝐴 / 𝑥𝐵 = ∅)
76imaeq1d 5968 . . . . . 6 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (∅ “ 𝐴 / 𝑥V))
8 0ima 5986 . . . . . 6 (∅ “ 𝐴 / 𝑥V) = ∅
97, 8eqtrdi 2794 . . . . 5 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = ∅)
106imaeq1d 5968 . . . . 5 𝐴 ∈ V → (𝐴 / 𝑥𝐵 “ V) = (∅ “ V))
115, 9, 103eqtr4a 2804 . . . 4 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V))
123, 11pm2.61i 182 . . 3 (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V)
131, 12eqtri 2766 . 2 𝐴 / 𝑥(𝐵 “ V) = (𝐴 / 𝑥𝐵 “ V)
14 dfrn4 6105 . . 3 ran 𝐵 = (𝐵 “ V)
1514csbeq2i 3840 . 2 𝐴 / 𝑥ran 𝐵 = 𝐴 / 𝑥(𝐵 “ V)
16 dfrn4 6105 . 2 ran 𝐴 / 𝑥𝐵 = (𝐴 / 𝑥𝐵 “ V)
1713, 15, 163eqtr4i 2776 1 𝐴 / 𝑥ran 𝐵 = ran 𝐴 / 𝑥𝐵
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3   = wceq 1539  wcel 2106  Vcvv 3432  csb 3832  c0 4256  ran crn 5590  cima 5592
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-nfc 2889  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3434  df-sbc 3717  df-csb 3833  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-br 5075  df-opab 5137  df-xp 5595  df-rel 5596  df-cnv 5597  df-dm 5599  df-rn 5600  df-res 5601  df-ima 5602
This theorem is referenced by:  sbcfg  6598  csbima12gALTVD  42517
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