Theorem bj-ssbn 32641

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 2019, bj-ax12 32634. Compare sbn 2391. (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.

Step	Hyp	Ref	Expression
1		df-ssb 32620	. 2 ⊢ ([𝑡/𝑥]b ¬ 𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
2		alinexa 1770	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → ¬ 𝜑) ↔ ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
3	2	imbi2i 326	. . 3 ⊢ ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)) ↔ (𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
4	3	albii 1747	. 2 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
5		alinexa 1770	. . 3 ⊢ (∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ ¬ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
6		bj-dfssb2 32640	. . 3 ⊢ ([𝑡/𝑥]b𝜑 ↔ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
7	5, 6	xchbinxr 325	. 2 ⊢ (∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ ¬ [𝑡/𝑥]b𝜑)
8	1, 4, 7	3bitri 286	1 ⊢ ([𝑡/𝑥]b ¬ 𝜑 ↔ ¬ [𝑡/𝑥]b𝜑)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196   ∧ wa 384  ∀wal 1481  ∃wex 1704  [wssb 32619

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-ssb 32620

This theorem is referenced by: (None)