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Metamath Proof Explorer

Theorem sbi1

Description: Distribute substitution over implication. (Contributed by NM, 14-May-1993) Remove dependencies on axioms. (Revised by Steven Nguyen, 24-Jul-2023)

Ref Expression
Assertion sbi1 ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) → ( [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )

Proof

Step Hyp Ref Expression
1 dfsb ⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ∀ 𝑧 ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → ( 𝜑 → 𝜓 ) ) ) )
2 ax-2 ⊢ ( ( 𝑥 = 𝑧 → ( 𝜑 → 𝜓 ) ) → ( ( 𝑥 = 𝑧 → 𝜑 ) → ( 𝑥 = 𝑧 → 𝜓 ) ) )
3 2 al2imi ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑧 → ( 𝜑 → 𝜓 ) ) → ( ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜓 ) ) )
4 3 imim3i ⊢ ( ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → ( 𝜑 → 𝜓 ) ) ) → ( ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜑 ) ) → ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜓 ) ) ) )
5 4 al2imi ⊢ ( ∀ 𝑧 ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → ( 𝜑 → 𝜓 ) ) ) → ( ∀ 𝑧 ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜑 ) ) → ∀ 𝑧 ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜓 ) ) ) )
6 dfsb ⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑧 ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜑 ) ) )
7 dfsb ⊢ ( [ 𝑦 / 𝑥 ] 𝜓 ↔ ∀ 𝑧 ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜓 ) ) )
8 5 6 7 3imtr4g ⊢ ( ∀ 𝑧 ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → ( 𝜑 → 𝜓 ) ) ) → ( [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )
9 1 8 sylbi ⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) → ( [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )