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Theorem bj-ssbn 31830
 Description: The result of a substitution in the negation of a formula is the negation of the result of the same substitution in that formula. Proved from Tarski, ax-10 2006, bj-ax12 31823. Compare sbn 2379. (Contributed by BJ, 25-Dec-2020.)
Assertion
Ref Expression
bj-ssbn ([𝑡/𝑥]b ¬ 𝜑 ↔ ¬ [𝑡/𝑥]b𝜑)

Proof of Theorem bj-ssbn
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 df-ssb 31809 . 2 ([𝑡/𝑥]b ¬ 𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
2 alinexa 1759 . . . 4 (∀𝑥(𝑥 = 𝑦 → ¬ 𝜑) ↔ ¬ ∃𝑥(𝑥 = 𝑦𝜑))
32imbi2i 325 . . 3 ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)) ↔ (𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦𝜑)))
43albii 1737 . 2 (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦𝜑)))
5 alinexa 1759 . . 3 (∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ ¬ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
6 bj-dfssb2 31829 . . 3 ([𝑡/𝑥]b𝜑 ↔ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦𝜑)))
75, 6xchbinxr 324 . 2 (∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦𝜑)) ↔ ¬ [𝑡/𝑥]b𝜑)
81, 4, 73bitri 285 1 ([𝑡/𝑥]b ¬ 𝜑 ↔ ¬ [𝑡/𝑥]b𝜑)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195   ∧ wa 383  ∀wal 1473  ∃wex 1695  [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-10 2006  ax-12 2034 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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