Theorem dfsb1

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 . 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 . (Contributed by BJ, 9-Jul-2023) Revise df-sb . (Revised by Wolf Lammen, 29-Jul-2023) (New usage is discouraged.)

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

Proof

Step	Hyp	Ref	Expression
1		sbequ2	⊢ ( 𝑥 = 𝑦 → ( [ 𝑦 / 𝑥 ] 𝜑 → 𝜑 ) )
2	1	com12	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ( 𝑥 = 𝑦 → 𝜑 ) )
3		sb1	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) )
4	2 3	jca	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 → ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) )
5		id	⊢ ( 𝑥 = 𝑦 → 𝑥 = 𝑦 )
6		sbequ1	⊢ ( 𝑥 = 𝑦 → ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜑 ) )
7	5 6	embantd	⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 = 𝑦 → 𝜑 ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
8	7	sps	⊢ ( ∀ 𝑥 𝑥 = 𝑦 → ( ( 𝑥 = 𝑦 → 𝜑 ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
9	8	adantrd	⊢ ( ∀ 𝑥 𝑥 = 𝑦 → ( ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
10		sb3	⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
11	10	adantld	⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) → [ 𝑦 / 𝑥 ] 𝜑 ) )
12	9 11	pm2.61i	⊢ ( ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) → [ 𝑦 / 𝑥 ] 𝜑 )
13	4 12	impbii	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ( ( 𝑥 = 𝑦 → 𝜑 ) ∧ ∃ 𝑥 ( 𝑥 = 𝑦 ∧ 𝜑 ) ) )