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