Metamath Proof Explorer


Theorem mdandyvrx6

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 mdandyvrx6.1
|- ( ph \/_ ze )
mdandyvrx6.2
|- ( ps \/_ si )
mdandyvrx6.3
|- ( ch <-> ph )
mdandyvrx6.4
|- ( th <-> ps )
mdandyvrx6.5
|- ( ta <-> ps )
mdandyvrx6.6
|- ( et <-> ph )
Assertion mdandyvrx6
|- ( ( ( ( ch \/_ ze ) /\ ( th \/_ si ) ) /\ ( ta \/_ si ) ) /\ ( et \/_ ze ) )

Proof

Step Hyp Ref Expression
1 mdandyvrx6.1
 |-  ( ph \/_ ze )
2 mdandyvrx6.2
 |-  ( ps \/_ si )
3 mdandyvrx6.3
 |-  ( ch <-> ph )
4 mdandyvrx6.4
 |-  ( th <-> ps )
5 mdandyvrx6.5
 |-  ( ta <-> ps )
6 mdandyvrx6.6
 |-  ( et <-> ph )
7 1 3 axorbciffatcxorb
 |-  ( ch \/_ ze )
8 2 4 axorbciffatcxorb
 |-  ( th \/_ si )
9 7 8 pm3.2i
 |-  ( ( ch \/_ ze ) /\ ( th \/_ si ) )
10 2 5 axorbciffatcxorb
 |-  ( ta \/_ si )
11 9 10 pm3.2i
 |-  ( ( ( ch \/_ ze ) /\ ( th \/_ si ) ) /\ ( ta \/_ si ) )
12 1 6 axorbciffatcxorb
 |-  ( et \/_ ze )
13 11 12 pm3.2i
 |-  ( ( ( ( ch \/_ ze ) /\ ( th \/_ si ) ) /\ ( ta \/_ si ) ) /\ ( et \/_ ze ) )