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Theorem csbrn 6150
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 6028 . . 3 𝐴 / 𝑥(𝐵 “ V) = (𝐴 / 𝑥𝐵𝐴 / 𝑥V)
2 csbconstg 3869 . . . . 5 (𝐴 ∈ V → 𝐴 / 𝑥V = V)
32imaeq2d 6009 . . . 4 (𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V))
4 0ima 6027 . . . . . 6 (∅ “ V) = ∅
54eqcomi 2740 . . . . 5 ∅ = (∅ “ V)
6 csbprc 4359 . . . . . . 7 𝐴 ∈ V → 𝐴 / 𝑥𝐵 = ∅)
76imaeq1d 6008 . . . . . 6 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (∅ “ 𝐴 / 𝑥V))
8 0ima 6027 . . . . . 6 (∅ “ 𝐴 / 𝑥V) = ∅
97, 8eqtrdi 2782 . . . . 5 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = ∅)
106imaeq1d 6008 . . . . 5 𝐴 ∈ V → (𝐴 / 𝑥𝐵 “ V) = (∅ “ V))
115, 9, 103eqtr4a 2792 . . . 4 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V))
123, 11pm2.61i 182 . . 3 (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V)
131, 12eqtri 2754 . 2 𝐴 / 𝑥(𝐵 “ V) = (𝐴 / 𝑥𝐵 “ V)
14 dfrn4 6149 . . 3 ran 𝐵 = (𝐵 “ V)
1514csbeq2i 3858 . 2 𝐴 / 𝑥ran 𝐵 = 𝐴 / 𝑥(𝐵 “ V)
16 dfrn4 6149 . 2 ran 𝐴 / 𝑥𝐵 = (𝐴 / 𝑥𝐵 “ V)
1713, 15, 163eqtr4i 2764 1 𝐴 / 𝑥ran 𝐵 = ran 𝐴 / 𝑥𝐵
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3   = wceq 1541  wcel 2111  Vcvv 3436  csb 3850  c0 4283  ran crn 5617  cima 5619
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-10 2144  ax-11 2160  ax-12 2180  ax-ext 2703  ax-sep 5234  ax-nul 5244  ax-pr 5370
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-nfc 2881  df-ral 3048  df-rex 3057  df-rab 3396  df-v 3438  df-sbc 3742  df-csb 3851  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4284  df-if 4476  df-sn 4577  df-pr 4579  df-op 4583  df-br 5092  df-opab 5154  df-xp 5622  df-rel 5623  df-cnv 5624  df-dm 5626  df-rn 5627  df-res 5628  df-ima 5629
This theorem is referenced by:  sbcfg  6649  csbima12gALTVD  44928
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