Theorem bj-ssb0 31817

Description: Substitution for a variable not occurring in a proposition. See sbf 2368. (Contributed by BJ, 22-Dec-2020.)

Assertion

Ref	Expression
bj-ssb0	⊢ ([𝑡/𝑥]b𝜑 ↔ 𝜑)

Distinct variable group:   𝜑,𝑥

Allowed substitution hint:   𝜑(𝑡)

Proof of Theorem bj-ssb0

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-ssb 31809	. 2 ⊢ ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
2		19.23v 1889	. . . . . . 7 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ (∃𝑥 𝑥 = 𝑦 → 𝜑))
3		ax6ev 1877	. . . . . . . . 9 ⊢ ∃𝑥 𝑥 = 𝑦
4		pm2.27 41	. . . . . . . . 9 ⊢ (∃𝑥 𝑥 = 𝑦 → ((∃𝑥 𝑥 = 𝑦 → 𝜑) → 𝜑))
5	3, 4	ax-mp 5	. . . . . . . 8 ⊢ ((∃𝑥 𝑥 = 𝑦 → 𝜑) → 𝜑)
6		ax-1 6	. . . . . . . 8 ⊢ (𝜑 → (∃𝑥 𝑥 = 𝑦 → 𝜑))
7	5, 6	impbii 198	. . . . . . 7 ⊢ ((∃𝑥 𝑥 = 𝑦 → 𝜑) ↔ 𝜑)
8	2, 7	bitri 263	. . . . . 6 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜑)
9	8	imbi2i 325	. . . . 5 ⊢ ((𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ (𝑦 = 𝑡 → 𝜑))
10	9	albii 1737	. . . 4 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → 𝜑))
11		19.23v 1889	. . . 4 ⊢ (∀𝑦(𝑦 = 𝑡 → 𝜑) ↔ (∃𝑦 𝑦 = 𝑡 → 𝜑))
12	10, 11	bitri 263	. . 3 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ (∃𝑦 𝑦 = 𝑡 → 𝜑))
13		ax6ev 1877	. . . . 5 ⊢ ∃𝑦 𝑦 = 𝑡
14		pm2.27 41	. . . . 5 ⊢ (∃𝑦 𝑦 = 𝑡 → ((∃𝑦 𝑦 = 𝑡 → 𝜑) → 𝜑))
15	13, 14	ax-mp 5	. . . 4 ⊢ ((∃𝑦 𝑦 = 𝑡 → 𝜑) → 𝜑)
16		ax-1 6	. . . 4 ⊢ (𝜑 → (∃𝑦 𝑦 = 𝑡 → 𝜑))
17	15, 16	impbii 198	. . 3 ⊢ ((∃𝑦 𝑦 = 𝑡 → 𝜑) ↔ 𝜑)
18	12, 17	bitri 263	. 2 ⊢ (∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ 𝜑)
19	1, 18	bitri 263	1 ⊢ ([𝑡/𝑥]b𝜑 ↔ 𝜑)

Syntax hints:   → wi 4   ↔ wb 195  ∀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

This theorem depends on definitions:  df-bi 196  df-ex 1696  df-ssb 31809

This theorem is referenced by: (None)