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Theorem bj-ssbequ1 33150
Description: This uses ax-12 2213 with a direct reference to ax12v 2214. Therefore, compared to bj-ax12 33141, there is a hidden use of sp 2217. Note that with ax-12 2213, it can be proved with disjoint variable condition on 𝑥, 𝑡. See sbequ1 2277. (Contributed by BJ, 22-Dec-2020.)
Assertion
Ref Expression
bj-ssbequ1 (𝑥 = 𝑡 → (𝜑 → [𝑡/𝑥]b𝜑))

Proof of Theorem bj-ssbequ1
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 equtr2 2126 . . . . . . . 8 ((𝑦 = 𝑡𝑥 = 𝑡) → 𝑦 = 𝑥)
21equcomd 2118 . . . . . . 7 ((𝑦 = 𝑡𝑥 = 𝑡) → 𝑥 = 𝑦)
3 ax12v 2214 . . . . . . 7 (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
42, 3syl 17 . . . . . 6 ((𝑦 = 𝑡𝑥 = 𝑡) → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
54expimpd 446 . . . . 5 (𝑦 = 𝑡 → ((𝑥 = 𝑡𝜑) → ∀𝑥(𝑥 = 𝑦𝜑)))
65com12 32 . . . 4 ((𝑥 = 𝑡𝜑) → (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
76alrimiv 2023 . . 3 ((𝑥 = 𝑡𝜑) → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
87ex 402 . 2 (𝑥 = 𝑡 → (𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑))))
9 df-ssb 33127 . 2 ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦𝜑)))
108, 9syl6ibr 244 1 (𝑥 = 𝑡 → (𝜑 → [𝑡/𝑥]b𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 385  wal 1651  [wssb 33126
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-12 2213
This theorem depends on definitions:  df-bi 199  df-an 386  df-ex 1876  df-ssb 33127
This theorem is referenced by:  bj-ssbid1  33153
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