Theorem 3mix1 1197

Description: Introduction in triple disjunction. (Contributed by NM, 4-Apr-1995.)

Assertion

Ref	Expression
3mix1	⊢ (𝜑 → (𝜑 ∨ 𝜓 ∨ 𝜒))

Proof of Theorem 3mix1

Step	Hyp	Ref	Expression
1		orc 724	. 2 ⊢ (𝜑 → (𝜑 ∨ (𝜓 ∨ 𝜒)))
2		3orass 1012	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒)))
3	1, 2	sylibr 134	1 ⊢ (𝜑 → (𝜑 ∨ 𝜓 ∨ 𝜒))

This proof depends on syntax axioms:   → wi 4   ∨ wo 720   ∨ w3o 1008

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721

This proof depends on definitions:  df-bi 117  df-3or 1010

This theorem is used by:  3mix2  1198  3mix3  1199  3mix1i  1200  3mix1d  1203  3jaob  1343  nntri3or  6766  exmidontriimlem3  7580  elnn0z  9662  nn0le2is012  9733  nn01to3  10027  fztri3or  10454  zabsle1  16284