Theorem sbor

Description: Disjunction inside and outside of a substitution are equivalent. (Contributed by NM, 29-Sep-2002)

	Ref	Expression
Assertion	sbor	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∨ [ 𝑦 / 𝑥 ] 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		sbim	⊢ ( [ 𝑦 / 𝑥 ] ( ¬ 𝜑 → 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] ¬ 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )
2		sbn	⊢ ( [ 𝑦 / 𝑥 ] ¬ 𝜑 ↔ ¬ [ 𝑦 / 𝑥 ] 𝜑 )
3	2	imbi1i	⊢ ( ( [ 𝑦 / 𝑥 ] ¬ 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( ¬ [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )
4	1 3	bitri	⊢ ( [ 𝑦 / 𝑥 ] ( ¬ 𝜑 → 𝜓 ) ↔ ( ¬ [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )
5		df-or	⊢ ( ( 𝜑 ∨ 𝜓 ) ↔ ( ¬ 𝜑 → 𝜓 ) )
6	5	sbbii	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ [ 𝑦 / 𝑥 ] ( ¬ 𝜑 → 𝜓 ) )
7		df-or	⊢ ( ( [ 𝑦 / 𝑥 ] 𝜑 ∨ [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( ¬ [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )
8	4 6 7	3bitr4i	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∨ [ 𝑦 / 𝑥 ] 𝜓 ) )