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Metamath Proof Explorer

Theorem sbnf

Description: Move nonfree predicate in and out of substitution; see sbal and sbex . (Contributed by BJ, 2-May-2019) (Proof shortened by Wolf Lammen, 2-May-2025)

Ref Expression
Assertion sbnf ( [ 𝑧 / 𝑦 ] Ⅎ 𝑥 𝜑 ↔ Ⅎ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 )

Proof

Step Hyp Ref Expression
1 df-nf ⊢ ( Ⅎ 𝑥 𝜑 ↔ ( ∃ 𝑥 𝜑 → ∀ 𝑥 𝜑 ) )
2 1 sbbii ⊢ ( [ 𝑧 / 𝑦 ] Ⅎ 𝑥 𝜑 ↔ [ 𝑧 / 𝑦 ] ( ∃ 𝑥 𝜑 → ∀ 𝑥 𝜑 ) )
3 sbim ⊢ ( [ 𝑧 / 𝑦 ] ( ∃ 𝑥 𝜑 → ∀ 𝑥 𝜑 ) ↔ ( [ 𝑧 / 𝑦 ] ∃ 𝑥 𝜑 → [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ) )
4 sbex ⊢ ( [ 𝑧 / 𝑦 ] ∃ 𝑥 𝜑 ↔ ∃ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 )
5 sbal ⊢ ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 )
6 4 5 imbi12i ⊢ ( ( [ 𝑧 / 𝑦 ] ∃ 𝑥 𝜑 → [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ) ↔ ( ∃ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 → ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ) )
7 df-nf ⊢ ( Ⅎ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ↔ ( ∃ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 → ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ) )
8 6 7 bitr4i ⊢ ( ( [ 𝑧 / 𝑦 ] ∃ 𝑥 𝜑 → [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ) ↔ Ⅎ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 )
9 2 3 8 3bitri ⊢ ( [ 𝑧 / 𝑦 ] Ⅎ 𝑥 𝜑 ↔ Ⅎ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 )