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Metamath Proof Explorer

Theorem sbc3or

Description: sbcor with a 3-disjuncts. This proof is sbc3orgVD automatically translated and minimized. (Contributed by Alan Sare, 31-Dec-2011) (Revised by NM, 24-Aug-2018) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion sbc3or ( [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ∨ [ 𝐴 / 𝑥 ] 𝜒 ) )

Proof

Step Hyp Ref Expression
1 sbcor ⊢ ( [ 𝐴 / 𝑥 ] ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ∨ [ 𝐴 / 𝑥 ] 𝜒 ) )
2 df-3or ⊢ ( ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) )
3 2 bicomi ⊢ ( ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) ↔ ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) )
4 3 sbcbii ⊢ ( [ 𝐴 / 𝑥 ] ( ( 𝜑 ∨ 𝜓 ) ∨ 𝜒 ) ↔ [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) )
5 sbcor ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ) )
6 5 orbi1i ⊢ ( ( [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ) ∨ [ 𝐴 / 𝑥 ] 𝜒 ) ↔ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ) ∨ [ 𝐴 / 𝑥 ] 𝜒 ) )
7 1 4 6 3bitr3i ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ) ∨ [ 𝐴 / 𝑥 ] 𝜒 ) )
8 df-3or ⊢ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ∨ [ 𝐴 / 𝑥 ] 𝜒 ) ↔ ( ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ) ∨ [ 𝐴 / 𝑥 ] 𝜒 ) )
9 7 8 bitr4i ⊢ ( [ 𝐴 / 𝑥 ] ( 𝜑 ∨ 𝜓 ∨ 𝜒 ) ↔ ( [ 𝐴 / 𝑥 ] 𝜑 ∨ [ 𝐴 / 𝑥 ] 𝜓 ∨ [ 𝐴 / 𝑥 ] 𝜒 ) )