Theorem dfsb1 2480

Description: Alternate definition of substitution. Remark 9.1 in [Megill] p. 447 (p. 15 of the preprint). This was the original definition before df-sb 2068. Note that it does not require dummy variables in its definiens; this is done by having 𝑥 free in the first conjunct and bound in the second. Usage of this theorem is discouraged because it depends on ax-13 2371. (Contributed by BJ, 9-Jul-2023.) Revise df-sb 2068. (Revised by Wolf Lammen, 29-Jul-2023.) (New usage is discouraged.)

Assertion

Ref	Expression
dfsb1	⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))

Proof of Theorem dfsb1

Step	Hyp	Ref	Expression
1		sbequ2 2241	. . . 4 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → 𝜑))
2	1	com12 32	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → (𝑥 = 𝑦 → 𝜑))
3		sb1 2477	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
4	2, 3	jca 512	. 2 ⊢ ([𝑦 / 𝑥]𝜑 → ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
5		id 22	. . . . . 6 ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑦)
6		sbequ1 2240	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
7	5, 6	embantd 59	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑))
8	7	sps 2178	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑))
9	8	adantrd 492	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → [𝑦 / 𝑥]𝜑))
10		sb3 2476	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑))
11	10	adantld 491	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → [𝑦 / 𝑥]𝜑))
12	9, 11	pm2.61i 182	. 2 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → [𝑦 / 𝑥]𝜑)
13	4, 12	impbii 208	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1539  ∃wex 1781  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-10 2137  ax-12 2171  ax-13 2371

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-ex 1782  df-nf 1786  df-sb 2068

This theorem is referenced by:  drsb1  2494  bj-dfsb2  35704  frege55b  42633