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Theorem csbrn 6203
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 6079 . . 3 𝐴 / 𝑥(𝐵 “ V) = (𝐴 / 𝑥𝐵𝐴 / 𝑥V)
2 csbconstg 3869 . . . . 5 (𝐴 ∈ V → 𝐴 / 𝑥V = V)
32imaeq2d 6060 . . . 4 (𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V))
4 0ima 6078 . . . . . 6 (∅ “ V) = ∅
54eqcomi 2771 . . . . 5 ∅ = (∅ “ V)
6 csbprc 4370 . . . . . . 7 𝐴 ∈ V → 𝐴 / 𝑥𝐵 = ∅)
76imaeq1d 6059 . . . . . 6 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (∅ “ 𝐴 / 𝑥V))
8 0ima 6078 . . . . . 6 (∅ “ 𝐴 / 𝑥V) = ∅
97, 8eqtrdi 2813 . . . . 5 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = ∅)
106imaeq1d 6059 . . . . 5 𝐴 ∈ V → (𝐴 / 𝑥𝐵 “ V) = (∅ “ V))
115, 9, 103eqtr4a 2823 . . . 4 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V))
123, 11pm2.61i 184 . . 3 (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V)
131, 12eqtri 2785 . 2 𝐴 / 𝑥(𝐵 “ V) = (𝐴 / 𝑥𝐵 “ V)
14 dfrn4 6200 . . 3 ran 𝐵 = (𝐵 “ V)
1514csbeq2i 3858 . 2 𝐴 / 𝑥ran 𝐵 = 𝐴 / 𝑥(𝐵 “ V)
16 dfrn4 6200 . 2 ran 𝐴 / 𝑥𝐵 = (𝐴 / 𝑥𝐵 “ V)
1713, 15, 163eqtr4i 2795 1 𝐴 / 𝑥ran 𝐵 = ran 𝐴 / 𝑥𝐵
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   = wceq 1570  wcel 2145  Vcvv 3453  csb 3850  c0 4282  ran crn 5660  cima 5662
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734  ax-sep 5255  ax-pr 5402
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-sbc 3743  df-csb 3851  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-br 5108  df-opab 5172  df-xp 5665  df-rel 5666  df-cnv 5667  df-dm 5669  df-rn 5670  df-res 5671  df-ima 5672
This theorem is used by:  sbcfg  6704  csbima12gALTVD  45727
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