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Theorem eqsbc1 3002
Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2281. (Contributed by Andrew Salmon, 29-Jun-2011.)
Assertion
Ref Expression
eqsbc1 (𝐴𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵𝐴 = 𝐵))
Distinct variable group:   𝑥,𝐵
Allowed substitution hints:   𝐴(𝑥)   𝑉(𝑥)

Proof of Theorem eqsbc1
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 dfsbcq 2964 . 2 (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝑥 = 𝐵[𝐴 / 𝑥]𝑥 = 𝐵))
2 eqeq1 2184 . 2 (𝑦 = 𝐴 → (𝑦 = 𝐵𝐴 = 𝐵))
3 sbsbc 2966 . . 3 ([𝑦 / 𝑥]𝑥 = 𝐵[𝑦 / 𝑥]𝑥 = 𝐵)
4 eqsb1 2281 . . 3 ([𝑦 / 𝑥]𝑥 = 𝐵𝑦 = 𝐵)
53, 4bitr3i 186 . 2 ([𝑦 / 𝑥]𝑥 = 𝐵𝑦 = 𝐵)
61, 2, 5vtoclbg 2798 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵𝐴 = 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105   = wceq 1353  [wsb 1762  wcel 2148  [wsbc 2962
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-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2739  df-sbc 2963
This theorem is referenced by:  sbceqal  3018  eqsbc2  3023
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