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Theorem sbequ1 2240
Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.) Revise df-sb 2063. (Revised by BJ, 22-Dec-2020.)
Assertion
Ref Expression
sbequ1 (𝑥 = 𝑡 → (𝜑 → [𝑡 / 𝑥]𝜑))

Proof of Theorem sbequ1
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 equeucl 2024 . . . . 5 (𝑥 = 𝑡 → (𝑦 = 𝑡𝑥 = 𝑦))
2 ax12v 2168 . . . . 5 (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
31, 2syl6 35 . . . 4 (𝑥 = 𝑡 → (𝑦 = 𝑡 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑))))
43com23 86 . . 3 (𝑥 = 𝑡 → (𝜑 → (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑))))
54alrimdv 1923 . 2 (𝑥 = 𝑡 → (𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑))))
6 df-sb 2063 . 2 ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
75, 6syl6ibr 253 1 (𝑥 = 𝑡 → (𝜑 → [𝑡 / 𝑥]𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wal 1528  [wsb 2062
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-12 2167
This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1774  df-sb 2063
This theorem is referenced by:  sbequ12  2244  dfsb1  2504  dfsb2  2526  sbequiOLD  2528  sbi1OLD  2536  2eu6  2741  bj-ssbid1  33882  sb5ALT  40721  2pm13.193  40748  2pm13.193VD  41099  sb5ALTVD  41109
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