Theorem bj-ssbequ1 32769

Description: This uses ax-12 2087 with a direct reference to ax12v 2088. Therefore, compared to bj-ax12 32759, there is a hidden use of sp 2091. Note that with ax-12 2087, it can be proved with dv condition on 𝑥, 𝑡. See sbequ1 2148. (Contributed by BJ, 22-Dec-2020.)

Assertion

Ref	Expression
bj-ssbequ1	⊢ (𝑥 = 𝑡 → (𝜑 → [𝑡/𝑥]b𝜑))

Proof of Theorem bj-ssbequ1

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equtr2 2000	. . . . . . . 8 ⊢ ((𝑦 = 𝑡 ∧ 𝑥 = 𝑡) → 𝑦 = 𝑥)
2	1	equcomd 1992	. . . . . . 7 ⊢ ((𝑦 = 𝑡 ∧ 𝑥 = 𝑡) → 𝑥 = 𝑦)
3		ax12v 2088	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
4	2, 3	syl 17	. . . . . 6 ⊢ ((𝑦 = 𝑡 ∧ 𝑥 = 𝑡) → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
5	4	expimpd 628	. . . . 5 ⊢ (𝑦 = 𝑡 → ((𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
6	5	com12 32	. . . 4 ⊢ ((𝑥 = 𝑡 ∧ 𝜑) → (𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
7	6	alrimiv 1895	. . 3 ⊢ ((𝑥 = 𝑡 ∧ 𝜑) → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
8	7	ex 449	. 2 ⊢ (𝑥 = 𝑡 → (𝜑 → ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
9		df-ssb 32745	. 2 ⊢ ([𝑡/𝑥]b𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
10	8, 9	syl6ibr 242	1 ⊢ (𝑥 = 𝑡 → (𝜑 → [𝑡/𝑥]b𝜑))

Syntax hints:   → wi 4   ∧ wa 383  ∀wal 1521  [wssb 32744

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-12 2087

This theorem depends on definitions:  df-bi 197  df-an 385  df-ex 1745  df-ssb 32745

This theorem is referenced by:  bj-ssbid1  32772