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Theorem eqsbc1 3841
Description: Substitution for the left-hand side in an equality. Class version of eqsb1 2865. (Contributed by Andrew Salmon, 29-Jun-2011.) Avoid ax-13 2375. (Revised by Wolf Lammen, 29-Apr-2023.)
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 3793 . 2 (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝑥 = 𝐵[𝐴 / 𝑥]𝑥 = 𝐵))
2 eqeq1 2739 . 2 (𝑦 = 𝐴 → (𝑦 = 𝐵𝐴 = 𝐵))
3 sbsbc 3795 . . 3 ([𝑦 / 𝑥]𝑥 = 𝐵[𝑦 / 𝑥]𝑥 = 𝐵)
4 eqsb1 2865 . . 3 ([𝑦 / 𝑥]𝑥 = 𝐵𝑦 = 𝐵)
53, 4bitr3i 277 . 2 ([𝑦 / 𝑥]𝑥 = 𝐵𝑦 = 𝐵)
61, 2, 5vtoclbg 3557 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵𝐴 = 𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206   = wceq 1537  [wsb 2062  wcel 2106  [wsbc 3791
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-sbc 3792
This theorem is referenced by:  sbceqalOLD  3858  eqsbc2  3860  fmptsnd  7189  fvmptnn04if  22871  snfil  23888  f1omptsnlem  37319  mptsnunlem  37321  topdifinffinlem  37330  relowlpssretop  37347  iotavalb  44426  onfrALTlem5  44540  eqsbc2VD  44838  onfrALTlem5VD  44883
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