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 ) )