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Theorem bj-ssb1 32275
Description: A simplified definition of substitution in case of disjoint variables. See bj-ssb1a 32274 for the backward implication, which does not require ax-11 2031 (note that here, the version of ax-11 2031 with disjoint setvar metavariables would suffice). Compare sb6 2428. (Contributed by BJ, 22-Dec-2020.)
Assertion
Ref Expression
bj-ssb1 ([𝑡/𝑥]b𝜑 ↔ ∀𝑥(𝑥 = 𝑡𝜑))
Distinct variable group:   𝑥,𝑡
Allowed substitution hints:   𝜑(𝑥,𝑡)

Proof of Theorem bj-ssb1
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 19.21v 1865 . . 3 (∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
21albii 1744 . 2 (∀𝑦𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
3 19.23v 1899 . . . . 5 (∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)) ↔ (∃𝑦 𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
4 equequ2 1950 . . . . . . . 8 (𝑦 = 𝑡 → (𝑥 = 𝑦𝑥 = 𝑡))
54imbi1d 331 . . . . . . 7 (𝑦 = 𝑡 → ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑡𝜑)))
65pm5.74i 260 . . . . . 6 ((𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
76albii 1744 . . . . 5 (∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
8 ax6ev 1887 . . . . . 6 𝑦 𝑦 = 𝑡
98a1bi 352 . . . . 5 ((𝑥 = 𝑡𝜑) ↔ (∃𝑦 𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
103, 7, 93bitr4ri 293 . . . 4 ((𝑥 = 𝑡𝜑) ↔ ∀𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)))
1110albii 1744 . . 3 (∀𝑥(𝑥 = 𝑡𝜑) ↔ ∀𝑥𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)))
12 alcom 2034 . . 3 (∀𝑥𝑦(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ ∀𝑦𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)))
1311, 12bitri 264 . 2 (∀𝑥(𝑥 = 𝑡𝜑) ↔ ∀𝑦𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)))
14 df-ssb 32262 . 2 ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
152, 13, 143bitr4ri 293 1 ([𝑡/𝑥]b𝜑 ↔ ∀𝑥(𝑥 = 𝑡𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wal 1478  wex 1701  [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  ax-11 2031
This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1702  df-ssb 32262
This theorem is referenced by:  bj-ax12ssb  32277  bj-ssbssblem  32291  bj-ssbcom3lem  32292
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