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Theorem csbeq2 3148
Description: Substituting into equivalent classes gives equivalent results. (Contributed by Giovanni Mascellani, 9-Apr-2018.)
Assertion
Ref Expression
csbeq2 (∀𝑥 𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶)

Proof of Theorem csbeq2
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 eleq2 2293 . . . . 5 (𝐵 = 𝐶 → (𝑦𝐵𝑦𝐶))
21alimi 1501 . . . 4 (∀𝑥 𝐵 = 𝐶 → ∀𝑥(𝑦𝐵𝑦𝐶))
3 sbcbi2 3079 . . . 4 (∀𝑥(𝑦𝐵𝑦𝐶) → ([𝐴 / 𝑥]𝑦𝐵[𝐴 / 𝑥]𝑦𝐶))
42, 3syl 14 . . 3 (∀𝑥 𝐵 = 𝐶 → ([𝐴 / 𝑥]𝑦𝐵[𝐴 / 𝑥]𝑦𝐶))
54abbidv 2347 . 2 (∀𝑥 𝐵 = 𝐶 → {𝑦[𝐴 / 𝑥]𝑦𝐵} = {𝑦[𝐴 / 𝑥]𝑦𝐶})
6 df-csb 3125 . 2 𝐴 / 𝑥𝐵 = {𝑦[𝐴 / 𝑥]𝑦𝐵}
7 df-csb 3125 . 2 𝐴 / 𝑥𝐶 = {𝑦[𝐴 / 𝑥]𝑦𝐶}
85, 6, 73eqtr4g 2287 1 (∀𝑥 𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶)
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105  wal 1393   = wceq 1395  wcel 2200  {cab 2215  [wsbc 3028  csb 3124
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-11 1552  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211
This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-sbc 3029  df-csb 3125
This theorem is referenced by:  prodeq2w  12075
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