Theorem bj-eqs 37545

Description: A lemma for substitutions, proved from Tarski's FOL. The version without DV (𝑥, 𝑦) is true but requires ax-13 2402. The disjoint variable 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 1960	. 2 ⊢ (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))
3		exim 1867	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → (∃𝑥 𝑥 = 𝑦 → ∃𝑥𝜑))
4		ax6ev 2002	. . . 4 ⊢ ∃𝑥 𝑥 = 𝑦
5		pm2.27 43	. . . 4 ⊢ (∃𝑥 𝑥 = 𝑦 → ((∃𝑥 𝑥 = 𝑦 → ∃𝑥𝜑) → ∃𝑥𝜑))
6	4, 5	ax-mp 5	. . 3 ⊢ ((∃𝑥 𝑥 = 𝑦 → ∃𝑥𝜑) → ∃𝑥𝜑)
7		ax5e 1945	. . 3 ⊢ (∃𝑥𝜑 → 𝜑)
8	3, 6, 7	3syl 19	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → 𝜑)
9	2, 8	impbii 212	1 ⊢ (𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000

This proof depends on definitions:  df-bi 210  df-ex 1813

This theorem is used by:  bj-sb  37559