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Theorem eqsbc2 3023
Description: Substitution for the right-hand side in an equality. (Contributed by Alan Sare, 24-Oct-2011.) (Proof shortened by JJ, 7-Jul-2021.)
Assertion
Ref Expression
eqsbc2 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥𝐵 = 𝐴))
Distinct variable group:   𝑥,𝐵
Allowed substitution hints:   𝐴(𝑥)   𝑉(𝑥)

Proof of Theorem eqsbc2
StepHypRef Expression
1 eqsbc1 3002 . 2 (𝐴𝑉 → ([𝐴 / 𝑥]𝑥 = 𝐵𝐴 = 𝐵))
2 eqcom 2179 . . 3 (𝐵 = 𝑥𝑥 = 𝐵)
32sbcbii 3022 . 2 ([𝐴 / 𝑥]𝐵 = 𝑥[𝐴 / 𝑥]𝑥 = 𝐵)
4 eqcom 2179 . 2 (𝐵 = 𝐴𝐴 = 𝐵)
51, 3, 43bitr4g 223 1 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝑥𝐵 = 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 105   = wceq 1353  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: (None)
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