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Theorem bj-ssbcom3lem 33003
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 2114 . . . . . . 7 (𝑦 = 𝑡 → (𝑥 = 𝑦𝑥 = 𝑡))
21imbi1d 331 . . . . . 6 (𝑦 = 𝑡 → ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑡𝜑)))
32pm5.74i 261 . . . . 5 ((𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
43albii 1898 . . . 4 (∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ ∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)))
5 19.21v 2023 . . . 4 (∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
6 19.21v 2023 . . . 4 (∀𝑥(𝑦 = 𝑡 → (𝑥 = 𝑡𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
74, 5, 63bitr3i 291 . . 3 ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
87albii 1898 . 2 (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
9 bj-ssb1 32987 . . . 4 ([𝑦/𝑥]b𝜑 ↔ ∀𝑥(𝑥 = 𝑦𝜑))
109bj-ssbbii 32978 . . 3 ([𝑡/𝑦]b[𝑦/𝑥]b𝜑 ↔ [𝑡/𝑦]b𝑥(𝑥 = 𝑦𝜑))
11 bj-ssb1 32987 . . 3 ([𝑡/𝑦]b𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
1210, 11bitri 265 . 2 ([𝑡/𝑦]b[𝑦/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
13 bj-ssb1 32987 . . . 4 ([𝑡/𝑥]b𝜑 ↔ ∀𝑥(𝑥 = 𝑡𝜑))
1413bj-ssbbii 32978 . . 3 ([𝑡/𝑦]b[𝑡/𝑥]b𝜑 ↔ [𝑡/𝑦]b𝑥(𝑥 = 𝑡𝜑))
15 bj-ssb1 32987 . . 3 ([𝑡/𝑦]b𝑥(𝑥 = 𝑡𝜑) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
1614, 15bitri 265 . 2 ([𝑡/𝑦]b[𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑡𝜑)))
178, 12, 163bitr4i 293 1 ([𝑡/𝑦]b[𝑦/𝑥]b𝜑 ↔ [𝑡/𝑦]b[𝑡/𝑥]b𝜑)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 197  wal 1632  [wssb 32973
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1873  ax-4 1888  ax-5 1994  ax-6 2060  ax-7 2096  ax-11 2193
This theorem depends on definitions:  df-bi 198  df-an 384  df-ex 1856  df-ssb 32974
This theorem is referenced by: (None)
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