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 ] ( 𝜑𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) )

Proof

Step Hyp Ref Expression
1 sbcex ( [ 𝐴 / 𝑥 ] ( 𝜑𝜓 ) → 𝐴 ∈ V )
2 sbcex ( [ 𝐴 / 𝑥 ] 𝜑𝐴 ∈ V )
3 sbcex ( [ 𝐴 / 𝑥 ] 𝜓𝐴 ∈ V )
4 2 3 jaoi ( ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) → 𝐴 ∈ V )
5 dfsbcq2 ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] ( 𝜑𝜓 ) ↔ [ 𝐴 / 𝑥 ] ( 𝜑𝜓 ) ) )
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 ] ( 𝜑𝜓 ) ↔ ( [ 𝑦 / 𝑥 ] 𝜑 ∨ [ 𝑦 / 𝑥 ] 𝜓 ) )
10 5 8 9 vtoclbg ( 𝐴 ∈ V → ( [ 𝐴 / 𝑥 ] ( 𝜑𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) ) )
11 1 4 10 pm5.21nii ( [ 𝐴 / 𝑥 ] ( 𝜑𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜓 ) )