Theorem sbbi

Description: Equivalence inside and outside of a substitution are equivalent. (Contributed by NM, 14-May-1993)

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

Proof

Step	Hyp	Ref	Expression
1		dfbi2	⊢ ( ( 𝜑 ↔ 𝜓 ) ↔ ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) )
2	1	sbbii	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜓 ) ↔ [ 𝑦 / 𝑥 ] ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) )
3		sbim	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )
4		sbim	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜓 → 𝜑 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜓 → [ 𝑦 / 𝑥 ] 𝜑 ) )
5	3 4	anbi12i	⊢ ( ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ∧ [ 𝑦 / 𝑥 ] ( 𝜓 → 𝜑 ) ) ↔ ( ( [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) ∧ ( [ 𝑦 / 𝑥 ] 𝜓 → [ 𝑦 / 𝑥 ] 𝜑 ) ) )
6		sban	⊢ ( [ 𝑦 / 𝑥 ] ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) ↔ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ∧ [ 𝑦 / 𝑥 ] ( 𝜓 → 𝜑 ) ) )
7		dfbi2	⊢ ( ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( ( [ 𝑦 / 𝑥 ] 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) ∧ ( [ 𝑦 / 𝑥 ] 𝜓 → [ 𝑦 / 𝑥 ] 𝜑 ) ) )
8	5 6 7	3bitr4i	⊢ ( [ 𝑦 / 𝑥 ] ( ( 𝜑 → 𝜓 ) ∧ ( 𝜓 → 𝜑 ) ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜓 ) )
9	2 8	bitri	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ↔ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] 𝜓 ) )