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Theorem csbifg 3690
 Description: Distribute proper substitution through the conditional operator. (Contributed by NM, 24-Feb-2013.) (Revised by Mario Carneiro, 14-Nov-2016.)
Assertion
Ref Expression
csbifg (A V[A / x] if(φ, B, C) = if([̣A / xφ, [A / x]B, [A / x]C))

Proof of Theorem csbifg
Dummy variable y is distinct from all other variables.
StepHypRef Expression
1 csbeq1 3139 . . 3 (y = A[y / x] if(φ, B, C) = [A / x] if(φ, B, C))
2 dfsbcq2 3049 . . . 4 (y = A → ([y / x]φ ↔ [̣A / xφ))
3 csbeq1 3139 . . . 4 (y = A[y / x]B = [A / x]B)
4 csbeq1 3139 . . . 4 (y = A[y / x]C = [A / x]C)
52, 3, 4ifbieq12d 3684 . . 3 (y = A → if([y / x]φ, [y / x]B, [y / x]C) = if([̣A / xφ, [A / x]B, [A / x]C))
61, 5eqeq12d 2367 . 2 (y = A → ([y / x] if(φ, B, C) = if([y / x]φ, [y / x]B, [y / x]C) ↔ [A / x] if(φ, B, C) = if([̣A / xφ, [A / x]B, [A / x]C)))
7 vex 2862 . . 3 y V
8 nfs1v 2106 . . . 4 x[y / x]φ
9 nfcsb1v 3168 . . . 4 x[y / x]B
10 nfcsb1v 3168 . . . 4 x[y / x]C
118, 9, 10nfif 3686 . . 3 x if([y / x]φ, [y / x]B, [y / x]C)
12 sbequ12 1919 . . . 4 (x = y → (φ ↔ [y / x]φ))
13 csbeq1a 3144 . . . 4 (x = yB = [y / x]B)
14 csbeq1a 3144 . . . 4 (x = yC = [y / x]C)
1512, 13, 14ifbieq12d 3684 . . 3 (x = y → if(φ, B, C) = if([y / x]φ, [y / x]B, [y / x]C))
167, 11, 15csbief 3177 . 2 [y / x] if(φ, B, C) = if([y / x]φ, [y / x]B, [y / x]C)
176, 16vtoclg 2914 1 (A V[A / x] if(φ, B, C) = if([̣A / xφ, [A / x]B, [A / x]C))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1642  [wsb 1648   ∈ wcel 1710  [̣wsbc 3046  [csb 3136   ifcif 3662 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3an 936  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-rab 2623  df-v 2861  df-sbc 3047  df-csb 3137  df-nin 3211  df-compl 3212  df-un 3214  df-if 3663 This theorem is referenced by: (None)
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