Theorem sbal2

Description: Move quantifier in and out of substitution. Usage of this theorem is discouraged because it depends on ax-13 . Check out sbal for a version replacing the distinctor with a disjoint variable condition, requiring fewer axioms. (Contributed by NM, 2-Jan-2002) Remove a distinct variable constraint. (Revised by Wolf Lammen, 24-Dec-2022) (Proof shortened by Wolf Lammen, 23-Sep-2023) (New usage is discouraged.)

	Ref	Expression
Assertion	sbal2	⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ) )

Proof

Step	Hyp	Ref	Expression
1		sbequ12	⊢ ( 𝑦 = 𝑧 → ( ∀ 𝑥 𝜑 ↔ [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ) )
2	1	sps	⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ( ∀ 𝑥 𝜑 ↔ [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ) )
3		sbequ12	⊢ ( 𝑦 = 𝑧 → ( 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜑 ) )
4	3	sps	⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ( 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜑 ) )
5	4	dral2	⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ( ∀ 𝑥 𝜑 ↔ ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ) )
6	2 5	bitr3d	⊢ ( ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ) )
7	6	adantl	⊢ ( ( ¬ ∀ 𝑥 𝑥 = 𝑦 ∧ ∀ 𝑦 𝑦 = 𝑧 ) → ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ) )
8		sb4b	⊢ ( ¬ ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → ∀ 𝑥 𝜑 ) ) )
9	8	adantl	⊢ ( ( ¬ ∀ 𝑥 𝑥 = 𝑦 ∧ ¬ ∀ 𝑦 𝑦 = 𝑧 ) → ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → ∀ 𝑥 𝜑 ) ) )
10		nfnae	⊢ Ⅎ 𝑥 ¬ ∀ 𝑦 𝑦 = 𝑧
11		sb4b	⊢ ( ¬ ∀ 𝑦 𝑦 = 𝑧 → ( [ 𝑧 / 𝑦 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → 𝜑 ) ) )
12	10 11	albid	⊢ ( ¬ ∀ 𝑦 𝑦 = 𝑧 → ( ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 ∀ 𝑦 ( 𝑦 = 𝑧 → 𝜑 ) ) )
13		alcom	⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝑦 = 𝑧 → 𝜑 ) ↔ ∀ 𝑦 ∀ 𝑥 ( 𝑦 = 𝑧 → 𝜑 ) )
14	12 13	syl6bb	⊢ ( ¬ ∀ 𝑦 𝑦 = 𝑧 → ( ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ↔ ∀ 𝑦 ∀ 𝑥 ( 𝑦 = 𝑧 → 𝜑 ) ) )
15		nfnae	⊢ Ⅎ 𝑦 ¬ ∀ 𝑥 𝑥 = 𝑦
16		nfeqf1	⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → Ⅎ 𝑥 𝑦 = 𝑧 )
17		19.21t	⊢ ( Ⅎ 𝑥 𝑦 = 𝑧 → ( ∀ 𝑥 ( 𝑦 = 𝑧 → 𝜑 ) ↔ ( 𝑦 = 𝑧 → ∀ 𝑥 𝜑 ) ) )
18	16 17	syl	⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ∀ 𝑥 ( 𝑦 = 𝑧 → 𝜑 ) ↔ ( 𝑦 = 𝑧 → ∀ 𝑥 𝜑 ) ) )
19	15 18	albid	⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ∀ 𝑦 ∀ 𝑥 ( 𝑦 = 𝑧 → 𝜑 ) ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → ∀ 𝑥 𝜑 ) ) )
20	14 19	sylan9bbr	⊢ ( ( ¬ ∀ 𝑥 𝑥 = 𝑦 ∧ ¬ ∀ 𝑦 𝑦 = 𝑧 ) → ( ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ↔ ∀ 𝑦 ( 𝑦 = 𝑧 → ∀ 𝑥 𝜑 ) ) )
21	9 20	bitr4d	⊢ ( ( ¬ ∀ 𝑥 𝑥 = 𝑦 ∧ ¬ ∀ 𝑦 𝑦 = 𝑧 ) → ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ) )
22	7 21	pm2.61dan	⊢ ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ) )