Theorem bj-subst 37264

Description: Proof of sbalex 2278 from core axioms, ax-10 2176 (modal5), and bj-ax12 37260. (Contributed by BJ, 29-Dec-2020.) (Proof modification is discouraged.)

Assertion

Ref	Expression
bj-subst	⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem bj-subst

Step	Hyp	Ref	Expression
1		bj-ax12 37260	. . . 4 ⊢ ∀𝑥(𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
2		pm3.31 454	. . . . 5 ⊢ ((𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))) → ((𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
3	2	aleximi 1862	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))) → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥∀𝑥(𝑥 = 𝑦 → 𝜑)))
4	1, 3	ax-mp 5	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥∀𝑥(𝑥 = 𝑦 → 𝜑))
5		hbe1a 2179	. . 3 ⊢ (∃𝑥∀𝑥(𝑥 = 𝑦 → 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑))
6	4, 5	syl 18	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑦 → 𝜑))
7		equs4v 2030	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
8	6, 7	impbii 212	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400  ∀wal 1568  ∃wex 1809

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-10 2176  ax-12 2213

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1810

This theorem is referenced by: (None)