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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