Theorem bj-eqs 31850

Description: A lemma for substitutions, proved from Tarski's FOL. The version without DV(𝑥, 𝑦) is true but requires ax-13 2234. The DV condition DV( 𝑥, 𝜑) is necessary for both directions: consider substituting 𝑥 = 𝑧 for 𝜑. (Contributed by BJ, 25-May-2021.)

Assertion

Ref	Expression
bj-eqs	⊢ (𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Distinct variable groups:   𝑥,𝑦   𝜑,𝑥

Allowed substitution hint:   𝜑(𝑦)

Proof of Theorem bj-eqs

Step	Hyp	Ref	Expression
1		ax-1 6	. . 3 ⊢ (𝜑 → (𝑥 = 𝑦 → 𝜑))
2	1	alrimiv 1842	. 2 ⊢ (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))
3		exim 1751	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → (∃𝑥 𝑥 = 𝑦 → ∃𝑥𝜑))
4		ax6ev 1877	. . . 4 ⊢ ∃𝑥 𝑥 = 𝑦
5		pm2.27 41	. . . 4 ⊢ (∃𝑥 𝑥 = 𝑦 → ((∃𝑥 𝑥 = 𝑦 → ∃𝑥𝜑) → ∃𝑥𝜑))
6	4, 5	ax-mp 5	. . 3 ⊢ ((∃𝑥 𝑥 = 𝑦 → ∃𝑥𝜑) → ∃𝑥𝜑)
7		ax5e 1829	. . 3 ⊢ (∃𝑥𝜑 → 𝜑)
8	3, 6, 7	3syl 18	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → 𝜑)
9	2, 8	impbii 198	1 ⊢ (𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 195  ∀wal 1473  ∃wex 1695

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

This theorem is referenced by:  bj-sb  31864