Metamath Proof Explorer


Theorem mdandyvrx15

Description: Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly. (Contributed by Jarvin Udandy, 7-Sep-2016)

Ref Expression
Hypotheses mdandyvrx15.1
|- ( ph \/_ ze )
mdandyvrx15.2
|- ( ps \/_ si )
mdandyvrx15.3
|- ( ch <-> ps )
mdandyvrx15.4
|- ( th <-> ps )
mdandyvrx15.5
|- ( ta <-> ps )
mdandyvrx15.6
|- ( et <-> ps )
Assertion mdandyvrx15
|- ( ( ( ( ch \/_ si ) /\ ( th \/_ si ) ) /\ ( ta \/_ si ) ) /\ ( et \/_ si ) )

Proof

Step Hyp Ref Expression
1 mdandyvrx15.1
 |-  ( ph \/_ ze )
2 mdandyvrx15.2
 |-  ( ps \/_ si )
3 mdandyvrx15.3
 |-  ( ch <-> ps )
4 mdandyvrx15.4
 |-  ( th <-> ps )
5 mdandyvrx15.5
 |-  ( ta <-> ps )
6 mdandyvrx15.6
 |-  ( et <-> ps )
7 2 1 3 4 5 6 mdandyvrx0
 |-  ( ( ( ( ch \/_ si ) /\ ( th \/_ si ) ) /\ ( ta \/_ si ) ) /\ ( et \/_ si ) )