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		df-sb	⊢ ( [ 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		df-sb	⊢ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑧 ( 𝑧 = 𝑦 → ∀ 𝑥 ( 𝑥 = 𝑧 → 𝜑 ) ) )
7		df-sb	⊢ ( [ 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 ) )