Theorem bj-ssbequ 31818

Description: Equality property for substitution, from Tarski's system. Compare sbequ 2364. (Contributed by BJ, 30-Dec-2020.)

Assertion

Ref	Expression
bj-ssbequ	⊢ (𝑠 = 𝑡 → ([𝑠/𝑥]b𝜑 ↔ [𝑡/𝑥]b𝜑))

Proof of Theorem bj-ssbequ

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equequ2 1940	. . . 4 ⊢ (𝑠 = 𝑡 → (𝑦 = 𝑠 ↔ 𝑦 = 𝑡))
2	1	imbi1d 330	. . 3 ⊢ (𝑠 = 𝑡 → ((𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
3	2	albidv 1836	. 2 ⊢ (𝑠 = 𝑡 → (∀𝑦(𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦 → 𝜑)) ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
4		df-ssb 31809	. 2 ⊢ ([𝑠/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑠 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
5		df-ssb 31809	. 2 ⊢ ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
6	3, 4, 5	3bitr4g 302	1 ⊢ (𝑠 = 𝑡 → ([𝑠/𝑥]b𝜑 ↔ [𝑡/𝑥]b𝜑))

Syntax hints:   → wi 4   ↔ wb 195  ∀wal 1473  [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

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-ssb 31809

This theorem is referenced by: (None)