Iof proof for Lemma inconsistent-bool-eq2:
1. (inr
) = (inl )
2. case inr
of inl(x) => 0 | inr(x) => 1 = case inl of inl(x) => 0 | inr(x) => 1
False
by Reduce (-1)
1:
1: 2. 1 = 0
1: False
.| 11,40 |
) = (inl )
of inl(x) => 0 | inr(x) => 1 = case inl of inl(x) => 0 | inr(x) => 1
False
by Reduce (-1)
1:
1: 2. 1 = 0
1: False
.