Theorem dfsb1 2510

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 2070. 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 2390. (Contributed by BJ, 9-Jul-2023.) Revise df-sb 2070. (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 2250	. . . 4 ⊢ (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → 𝜑))
2	1	com12 32	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → (𝑥 = 𝑦 → 𝜑))
3		sb1 2503	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
4	2, 3	jca 514	. 2 ⊢ ([𝑦 / 𝑥]𝜑 → ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
5		id 22	. . . . . 6 ⊢ (𝑥 = 𝑦 → 𝑥 = 𝑦)
6		sbequ1 2249	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
7	5, 6	embantd 59	. . . . 5 ⊢ (𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑))
8	7	sps 2184	. . . 4 ⊢ (∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑))
9	8	adantrd 494	. . 3 ⊢ (∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → [𝑦 / 𝑥]𝜑))
10		sb3 2502	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑))
11	10	adantld 493	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → [𝑦 / 𝑥]𝜑))
12	9, 11	pm2.61i 184	. 2 ⊢ (((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → [𝑦 / 𝑥]𝜑)
13	4, 12	impbii 211	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ((𝑥 = 𝑦 → 𝜑) ∧ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398  ∀wal 1535  ∃wex 1780  [wsb 2069

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2145  ax-12 2177  ax-13 2390

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1781  df-nf 1785  df-sb 2070

This theorem is referenced by:  sb2vOLDOLD  2512  sbequ2OLDOLD  2514  sbimiOLD  2515  sbimdvOLD  2516  equsb1vOLDOLD  2517  sbimdOLD  2518  sbtvOLD  2519  sbequ1OLD  2520  drsb1  2535  sbnOLD  2541  subsym1  33775  bj-dfsb2  34161  frege55b  40292