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Theorem csbrn 6159
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 6035 . . 3 𝐴 / 𝑥(𝐵 “ V) = (𝐴 / 𝑥𝐵𝐴 / 𝑥V)
2 csbconstg 3878 . . . . 5 (𝐴 ∈ V → 𝐴 / 𝑥V = V)
32imaeq2d 6017 . . . 4 (𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V))
4 0ima 6034 . . . . . 6 (∅ “ V) = ∅
54eqcomi 2742 . . . . 5 ∅ = (∅ “ V)
6 csbprc 4370 . . . . . . 7 𝐴 ∈ V → 𝐴 / 𝑥𝐵 = ∅)
76imaeq1d 6016 . . . . . 6 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (∅ “ 𝐴 / 𝑥V))
8 0ima 6034 . . . . . 6 (∅ “ 𝐴 / 𝑥V) = ∅
97, 8eqtrdi 2789 . . . . 5 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = ∅)
106imaeq1d 6016 . . . . 5 𝐴 ∈ V → (𝐴 / 𝑥𝐵 “ V) = (∅ “ V))
115, 9, 103eqtr4a 2799 . . . 4 𝐴 ∈ V → (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V))
123, 11pm2.61i 182 . . 3 (𝐴 / 𝑥𝐵𝐴 / 𝑥V) = (𝐴 / 𝑥𝐵 “ V)
131, 12eqtri 2761 . 2 𝐴 / 𝑥(𝐵 “ V) = (𝐴 / 𝑥𝐵 “ V)
14 dfrn4 6158 . . 3 ran 𝐵 = (𝐵 “ V)
1514csbeq2i 3867 . 2 𝐴 / 𝑥ran 𝐵 = 𝐴 / 𝑥(𝐵 “ V)
16 dfrn4 6158 . 2 ran 𝐴 / 𝑥𝐵 = (𝐴 / 𝑥𝐵 “ V)
1713, 15, 163eqtr4i 2771 1 𝐴 / 𝑥ran 𝐵 = ran 𝐴 / 𝑥𝐵
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3   = wceq 1542  wcel 2107  Vcvv 3447  csb 3859  c0 4286  ran crn 5638  cima 5640
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 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2155  ax-12 2172  ax-ext 2704  ax-sep 5260  ax-nul 5267  ax-pr 5388
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-3an 1090  df-tru 1545  df-fal 1555  df-ex 1783  df-nf 1787  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-nfc 2886  df-ral 3062  df-rex 3071  df-rab 3407  df-v 3449  df-sbc 3744  df-csb 3860  df-dif 3917  df-un 3919  df-in 3921  df-ss 3931  df-nul 4287  df-if 4491  df-sn 4591  df-pr 4593  df-op 4597  df-br 5110  df-opab 5172  df-xp 5643  df-rel 5644  df-cnv 5645  df-dm 5647  df-rn 5648  df-res 5649  df-ima 5650
This theorem is referenced by:  sbcfg  6670  csbima12gALTVD  43271
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