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Theorem bj-ssbcom3lem 31839
Description: Lemma for bj-ssbcom3 when setvar variables are disjoint. Remark: does not seem useful. (Contributed by BJ, 30-Dec-2020.)
Assertion
Ref Expression
bj-ssbcom3lem ([𝑡/𝑦]b[𝑦/𝑥]b𝜑 ↔ [𝑡/𝑦]b[𝑡/𝑥]b𝜑)
Distinct variable group:   𝑥,𝑦,𝑡
Allowed substitution hints:   𝜑(𝑥,𝑦,𝑡)

Proof of Theorem bj-ssbcom3lem
StepHypRef Expression
1 equequ2 1940 . . . . . . 7 (𝑦 = 𝑡 → (𝑥 = 𝑦𝑥 = 𝑡))
21imbi1d 330 . . . . . 6 (𝑦 = 𝑡 → ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑡𝜑)))
32pm5.74i 259 . . . . 5 ((𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
43albii 1737 . . . 4 (∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ ∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
5 19.21v 1855 . . . 4 (∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
6 19.21v 1855 . . . 4 (∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
74, 5, 63bitr3i 289 . . 3 ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
87albii 1737 . 2 (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
9 bj-ssb1 31822 . . . 4 ([𝑦/𝑥]b𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
109bj-ssbbii 31813 . . 3 ([𝑡/𝑦]b[𝑦/𝑥]b𝜑 ↔ [𝑡/𝑦]b𝑥(𝑥 = 𝑦𝜑))
11 bj-ssb1 31822 . . 3 ([𝑡/𝑦]b𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
1210, 11bitri 263 . 2 ([𝑡/𝑦]b[𝑦/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
13 bj-ssb1 31822 . . . 4 ([𝑡/𝑥]b𝜑 ↔ ∀𝑥(𝑥 = 𝑡𝜑))
1413bj-ssbbii 31813 . . 3 ([𝑡/𝑦]b[𝑡/𝑥]b𝜑 ↔ [𝑡/𝑦]b𝑥(𝑥 = 𝑡𝜑))
15 bj-ssb1 31822 . . 3 ([𝑡/𝑦]b𝑥(𝑥 = 𝑡𝜑) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
1614, 15bitri 263 . 2 ([𝑡/𝑦]b[𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
178, 12, 163bitr4i 291 1 ([𝑡/𝑦]b[𝑦/𝑥]b𝜑 ↔ [𝑡/𝑦]b[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  ax-11 2021
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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