Metamath Proof Explorer


Theorem 3orbi123i

Description: Join 3 biconditionals with disjunction. (Contributed by NM, 17-May-1994)

Ref Expression
Hypotheses bi3.1
|- ( ph <-> ps )
bi3.2
|- ( ch <-> th )
bi3.3
|- ( ta <-> et )
Assertion 3orbi123i
|- ( ( ph \/ ch \/ ta ) <-> ( ps \/ th \/ et ) )

Proof

Step Hyp Ref Expression
1 bi3.1
 |-  ( ph <-> ps )
2 bi3.2
 |-  ( ch <-> th )
3 bi3.3
 |-  ( ta <-> et )
4 1 2 orbi12i
 |-  ( ( ph \/ ch ) <-> ( ps \/ th ) )
5 4 3 orbi12i
 |-  ( ( ( ph \/ ch ) \/ ta ) <-> ( ( ps \/ th ) \/ et ) )
6 df-3or
 |-  ( ( ph \/ ch \/ ta ) <-> ( ( ph \/ ch ) \/ ta ) )
7 df-3or
 |-  ( ( ps \/ th \/ et ) <-> ( ( ps \/ th ) \/ et ) )
8 5 6 7 3bitr4i
 |-  ( ( ph \/ ch \/ ta ) <-> ( ps \/ th \/ et ) )