Let x0 of type ο be given.
Let x1 of type ο be given.
Assume H0:
x1 ⟶ not x0.
Assume H1:
not x1 ⟶ x0.
Apply xm with
x1,
exactly1of2 x0 x1 leaving 2 subgoals.
Assume H2: x1.
Apply exactly1of2_I2 with
x0,
x1 leaving 2 subgoals.
Apply H0.
The subproof is completed by applying H2.
The subproof is completed by applying H2.
Apply exactly1of2_I1 with
x0,
x1 leaving 2 subgoals.
Apply H1.
The subproof is completed by applying H2.
The subproof is completed by applying H2.