Theorem 3mix3 1199

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

Assertion

Ref	Expression
3mix3	⊢ (𝜑 → (𝜓 ∨ 𝜒 ∨ 𝜑))

Proof of Theorem 3mix3

Step	Hyp	Ref	Expression
1		3mix1 1197	. 2 ⊢ (𝜑 → (𝜑 ∨ 𝜓 ∨ 𝜒))
2		3orrot 1015	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜒 ∨ 𝜑))
3	1, 2	sylib 122	1 ⊢ (𝜑 → (𝜓 ∨ 𝜒 ∨ 𝜑))

This proof depends on syntax axioms:   → wi 4   ∨ 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:  3mix3i  1202  3mix3d  1205  3jaob  1343  tpid3g  3828  funtpg  5432  exmidontriimlem3  7580  nn0le2is012  9733  nn01to3  10027  fztri3or  10454  qbtwnxr  10703  hashfiv01gt1  11237  pfxnd  11477  pfxwrdsymbg  11478