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Theorem sbequ1 2246
Description: An equality theorem for substitution. (Contributed by NM, 16-May-1993.) Revise df-sb 2070. (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 2031 . . . . 5 (𝑥 = 𝑡 → (𝑦 = 𝑡𝑥 = 𝑦))
2 ax12v 2176 . . . . 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 1930 . 2 (𝑥 = 𝑡 → (𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑))))
6 df-sb 2070 . 2 ([𝑡 / 𝑥]𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
75, 6syl6ibr 255 1 (𝑥 = 𝑡 → (𝜑 → [𝑡 / 𝑥]𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wal 1536  [wsb 2069
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-12 2175
This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-sb 2070
This theorem is referenced by:  sbequ12  2250  dfsb1  2499  dfsb2  2511  sbi1OLD  2519  2eu6  2719  bj-ssbid1  34110  sb5ALT  41231  2pm13.193  41258  2pm13.193VD  41609  sb5ALTVD  41619
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