Metamath Proof Explorer

Theorem sbc3an

Description: Distribution of class substitution over triple conjunction. (Contributed by NM, 14-Dec-2006) (Revised by NM, 17-Aug-2018)

Ref Expression
Assertion sbc3an ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )

Proof

Step Hyp Ref Expression
1 df-3an ⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) )
2 1 sbcbii ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ [ 𝐴 / 𝑥 ] ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) )
3 sbcan ⊢ ( [ 𝐴 / 𝑥 ] ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )
4 sbcan ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) )
5 4 anbi1i ⊢ ( ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ∧ [ 𝐴 / 𝑥 ] 𝜒 ) ↔ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )
6 2 3 5 3bitri ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )
7 df-3an ⊢ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ∧ [ 𝐴 / 𝑥 ] 𝜒 ) ↔ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )
8 6 7 bitr4i ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )