Let x0 of type ι be given.
Let x1 of type ι be given.
Let x2 of type ι be given.
Let x3 of type ι be given.
Let x4 of type ι be given.
Let x5 of type ι be given.
Apply beta with
6,
λ x6 . If_i (x6 = 0) x0 (If_i (x6 = 1) x1 (If_i (x6 = 2) x2 (If_i (x6 = 3) x3 (If_i (x6 = 4) x4 x5)))),
3,
λ x6 x7 . x7 = x3 leaving 2 subgoals.
The subproof is completed by applying In_3_6.
Apply If_i_0 with
3 = 0,
x0,
If_i (3 = 1) x1 (If_i (3 = 2) x2 (If_i (3 = 3) x3 (If_i (3 = 4) x4 x5))),
λ x6 x7 . x7 = x3 leaving 2 subgoals.
The subproof is completed by applying neq_3_0.
Apply If_i_0 with
3 = 1,
x1,
If_i (3 = 2) x2 (If_i (3 = 3) x3 (If_i (3 = 4) x4 x5)),
λ x6 x7 . x7 = x3 leaving 2 subgoals.
The subproof is completed by applying neq_3_1.
Apply If_i_0 with
3 = 2,
x2,
If_i (3 = 3) x3 (If_i (3 = 4) x4 x5),
λ x6 x7 . x7 = x3 leaving 2 subgoals.
The subproof is completed by applying neq_3_2.
Apply If_i_1 with
3 = 3,
x3,
If_i (3 = 4) x4 x5.
Let x6 of type ι → ι → ο be given.
Assume H0: x6 3 3.
The subproof is completed by applying H0.