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Metamath Proof Explorer

Theorem sb3an

Description: Threefold conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006)

Ref Expression
Assertion sb3an ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ∧ [ 𝑦 / 𝑥 ] 𝜒 ) )

Proof

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