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Proof step 4 from proof int2bits-leneq-neg (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
j < 0 ⊦ ¬ 2 ^ (n→i(# int2bits(j)) - 1) + j < 0, # int2bits(j) = 1
Proof step 12:
j < 0 ⊦ ¬ 0 ≤ j + 2 ^ (n→i(# int2bits(j)) - 2), # int2bits(j) = 1

Rule argument

pos: right 2

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