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Proof step 2 from proof int2bits-leneq-neg (previous step)

j < 0 ⊦ # int2bits(j) = n ↔ ¬ 2 ^ (n→i(n) - 1) + j < 0 ∧ ¬ 0 ≤ j + 2 ^ (n→i(n) - 2), n = 1, n = 0

User applied rule case distinction

And obtained the following new goals

Proof step 3:
# int2bits(j) = n, j < 0 ⊦ ¬ 2 ^ (n→i(n) - 1) + j < 0 ∧ ¬ 0 ≤ j + 2 ^ (n→i(n) - 2), n = 1, n = 0
Proof step 26:
¬ 2 ^ (n→i(n) - 1) + j < 0 ∧ ¬ 0 ≤ j + 2 ^ (n→i(n) - 2), j < 0 ⊦ # int2bits(j) = n, n = 1, n = 0

Rule argument

pos: right 1

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