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 unknownprop_c5e2164052a280ad5b04f622e53815f0267ee33361e4345305e43303abef2c1b 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)))),
1,
λ x6 x7 . x7 = x1 leaving 2 subgoals.
The subproof is completed by applying unknownprop_64ce2e061803b0344cfad1c28352ad525e84ee3ff63c0c512dbe3a51a5ac033c.
Apply unknownprop_5a150bd86f4285de5d98c60b17d4452a655b4d88de0a02247259cdad6e6d992c with
1 = 0,
x0,
If_i (1 = 1) x1 (If_i (1 = 2) x2 (If_i (1 = 3) x3 (If_i (1 = 4) x4 x5))),
λ x6 x7 . x7 = x1 leaving 2 subgoals.
The subproof is completed by applying unknownprop_698eb914d3aabc70ca0bb946b6907a27e3cce6e39040426b924e77df3507fbcf.
Apply unknownprop_6f44febdf8a865ee94133af873e3c2941a931de6ac80968301360290e02ca608 with
1 = 1,
x1,
If_i (1 = 2) x2 (If_i (1 = 3) x3 (If_i (1 = 4) x4 x5)).
Let x6 of type ι → ι → ο be given.
Assume H0: x6 1 1.
The subproof is completed by applying H0.