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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 sbcan ⊢ ( [ 𝐴 / 𝑥 ] ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )
2 sbcan ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) )
3 1 2 bianbi ⊢ ( [ 𝐴 / 𝑥 ] ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) ↔ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )
4 df-3an ⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( ( 𝜑 ∧ 𝜓 ) ∧ 𝜒 ) )
5 4 sbcbii ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ [ 𝐴 / 𝑥 ] ( ( 𝜑 ∧ 𝜓 ) ∧ 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 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ∧ [ 𝐴 / 𝑥 ] 𝜒 ) )