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Metamath Proof Explorer

Theorem sbal2

Description: Move quantifier in and out of substitution. (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) Usage of this theorem is discouraged because it depends on ax-13 . Use sbal instead. (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 bitrdi ⊢ ( ¬ ∀ 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 ) )