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Theorem bj-ssbequ 31818
 Description: Equality property for substitution, from Tarski's system. Compare sbequ 2364. (Contributed by BJ, 30-Dec-2020.)
Assertion
Ref Expression
bj-ssbequ (𝑠 = 𝑡 → ([𝑠/𝑥]b𝜑 ↔ [𝑡/𝑥]b𝜑))

Proof of Theorem bj-ssbequ
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 equequ2 1940 . . . 4 (𝑠 = 𝑡 → (𝑦 = 𝑠𝑦 = 𝑡))
21imbi1d 330 . . 3 (𝑠 = 𝑡 → ((𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑))))
32albidv 1836 . 2 (𝑠 = 𝑡 → (∀𝑦(𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑))))
4 df-ssb 31809 . 2 ([𝑠/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦𝜑)))
5 df-ssb 31809 . 2 ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
63, 4, 53bitr4g 302 1 (𝑠 = 𝑡 → ([𝑠/𝑥]b𝜑 ↔ [𝑡/𝑥]b𝜑))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195  ∀wal 1473  [wssb 31808 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922 This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-ssb 31809 This theorem is referenced by: (None)
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