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Metamath Proof Explorer

Theorem sbcan

Description: Distribution of class substitution over conjunction. (Contributed by NM, 31-Dec-2016) (Revised by NM, 17-Aug-2018)

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

Proof

Step Hyp Ref Expression
1 sbcex ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) → 𝐴 ∈ V )
2 sbcex ⊢ ( [ 𝐴 / 𝑥 ] 𝜓 → 𝐴 ∈ V )
3 2 adantl ⊢ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) → 𝐴 ∈ V )
4 dfsbcq2 ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ) )
5 dfsbcq2 ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝐴 / 𝑥 ] 𝜑 ) )
6 dfsbcq2 ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ [ 𝐴 / 𝑥 ] 𝜓 ) )
7 5 6 anbi12d ⊢ ( 𝑦 = 𝐴 → ( ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ) )
8 sban ⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∧ [ 𝑦 / 𝑥 ] 𝜓 ) )
9 4 7 8 vtoclbg ⊢ ( 𝐴 ∈ V → ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) ) )
10 1 3 9 pm5.21nii ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∧ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∧ [ 𝐴 / 𝑥 ] 𝜓 ) )