Metamath Proof Explorer


Theorem mdandyvrx1

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 mdandyvrx1.1
|- ( ph \/_ ze )
mdandyvrx1.2
|- ( ps \/_ si )
mdandyvrx1.3
|- ( ch <-> ps )
mdandyvrx1.4
|- ( th <-> ph )
mdandyvrx1.5
|- ( ta <-> ph )
mdandyvrx1.6
|- ( et <-> ph )
Assertion mdandyvrx1
|- ( ( ( ( ch \/_ si ) /\ ( th \/_ ze ) ) /\ ( ta \/_ ze ) ) /\ ( et \/_ ze ) )

Proof

Step Hyp Ref Expression
1 mdandyvrx1.1
 |-  ( ph \/_ ze )
2 mdandyvrx1.2
 |-  ( ps \/_ si )
3 mdandyvrx1.3
 |-  ( ch <-> ps )
4 mdandyvrx1.4
 |-  ( th <-> ph )
5 mdandyvrx1.5
 |-  ( ta <-> ph )
6 mdandyvrx1.6
 |-  ( et <-> ph )
7 2 3 axorbciffatcxorb
 |-  ( ch \/_ si )
8 1 4 axorbciffatcxorb
 |-  ( th \/_ ze )
9 7 8 pm3.2i
 |-  ( ( ch \/_ si ) /\ ( th \/_ ze ) )
10 1 5 axorbciffatcxorb
 |-  ( ta \/_ ze )
11 9 10 pm3.2i
 |-  ( ( ( ch \/_ si ) /\ ( th \/_ ze ) ) /\ ( ta \/_ ze ) )
12 1 6 axorbciffatcxorb
 |-  ( et \/_ ze )
13 11 12 pm3.2i
 |-  ( ( ( ( ch \/_ si ) /\ ( th \/_ ze ) ) /\ ( ta \/_ ze ) ) /\ ( et \/_ ze ) )