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Metamath Proof Explorer

Theorem sbcor

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

Ref Expression
Assertion sbcor ( [ 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 ] 𝜑 → 𝐴 ∈ V )
3 sbcex ⊢ ( [ 𝐴 / 𝑥 ] 𝜓 → 𝐴 ∈ V )
4 2 3 jaoi ⊢ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ) → 𝐴 ∈ V )
5 dfsbcq2 ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ) )
6 dfsbcq2 ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝐴 / 𝑥 ] 𝜑 ) )
7 dfsbcq2 ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜓 ↔ [ 𝐴 / 𝑥 ] 𝜓 ) )
8 6 7 orbi12d ⊢ ( 𝑦 = 𝐴 → ( ( [ 𝑦 / 𝑥 ] 𝜑 ∨ [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ) ) )
9 sbor ⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∨ [ 𝑦 / 𝑥 ] 𝜓 ) )
10 5 8 9 vtoclbg ⊢ ( 𝐴 ∈ V → ( [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ) ) )
11 1 4 10 pm5.21nii ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ) )