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 ] ( 𝜑𝜓𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓[ 𝐴 / 𝑥 ] 𝜒 ) )

Proof

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