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