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Theorem bj-ssbequ 32271
Description: Equality property for substitution, from Tarski's system. Compare sbequ 2375. (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 1950 . . . 4 (𝑠 = 𝑡 → (𝑦 = 𝑠𝑦 = 𝑡))
21imbi1d 331 . . 3 (𝑠 = 𝑡 → ((𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑))))
32albidv 1846 . 2 (𝑠 = 𝑡 → (∀𝑦(𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑))))
4 df-ssb 32262 . 2 ([𝑠/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦𝜑)))
5 df-ssb 32262 . 2 ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
63, 4, 53bitr4g 303 1 (𝑠 = 𝑡 → ([𝑠/𝑥]b𝜑 ↔ [𝑡/𝑥]b𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wal 1478  [wssb 32261
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932
This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1702  df-ssb 32262
This theorem is referenced by: (None)
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