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Theorem csbeq2 3108
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 2260 . . . . 5 (𝐵 = 𝐶 → (𝑦𝐵𝑦𝐶))
21alimi 1469 . . . 4 (∀𝑥 𝐵 = 𝐶 → ∀𝑥(𝑦𝐵𝑦𝐶))
3 sbcbi2 3040 . . . 4 (∀𝑥(𝑦𝐵𝑦𝐶) → ([𝐴 / 𝑥]𝑦𝐵[𝐴 / 𝑥]𝑦𝐶))
42, 3syl 14 . . 3 (∀𝑥 𝐵 = 𝐶 → ([𝐴 / 𝑥]𝑦𝐵[𝐴 / 𝑥]𝑦𝐶))
54abbidv 2314 . 2 (∀𝑥 𝐵 = 𝐶 → {𝑦[𝐴 / 𝑥]𝑦𝐵} = {𝑦[𝐴 / 𝑥]𝑦𝐶})
6 df-csb 3085 . 2 𝐴 / 𝑥𝐵 = {𝑦[𝐴 / 𝑥]𝑦𝐵}
7 df-csb 3085 . 2 𝐴 / 𝑥𝐶 = {𝑦[𝐴 / 𝑥]𝑦𝐶}
85, 6, 73eqtr4g 2254 1 (∀𝑥 𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶)
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105  wal 1362   = wceq 1364  wcel 2167  {cab 2182  [wsbc 2989  csb 3084
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 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-11 1520  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178
This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-sbc 2990  df-csb 3085
This theorem is referenced by:  prodeq2w  11721
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