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.

Step	Hyp	Ref	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 = 𝑦 ∧ 𝜑))
3	2	imbi2i 325	. . 3 ⊢ ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)) ↔ (𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
4	3	albii 1737	. 2 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
5		alinexa 1759	. . 3 ⊢ (∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ ¬ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
6		bj-dfssb2 31829	. . 3 ⊢ ([𝑡/𝑥]b𝜑 ↔ ∃𝑦(𝑦 = 𝑡 ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
7	5, 6	xchbinxr 324	. 2 ⊢ (∀𝑦(𝑦 = 𝑡 → ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ↔ ¬ [𝑡/𝑥]b𝜑)
8	1, 4, 7	3bitri 285	1 ⊢ ([𝑡/𝑥]b ¬ 𝜑 ↔ ¬ [𝑡/𝑥]b𝜑)

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)