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Proof step 24 from proof int2int-lastn-negpos (previous step)

# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), bits2int(bits) * 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))) = i, bits.first, i < 0, m ≤ # int2bits(i) ⊦ bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), i % 2 ^ n→i(m) = 0, m = 0, bits = [], lastn(m, int2bits(i)).first, ¬ bits2int(lastn(m, int2bits(i))) < 0 ∧ bits2int(lastn(m, int2bits(i))) % 2 ^ n→i(m) ≠ 0

User applied rule case distinction

And obtained the following new goals

Proof step 25:
# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), bits2int(bits) * 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))) = i, bits.first, i < 0, m ≤ # int2bits(i) ⊦ bits2int(lastn(m, int2bits(i))) % 2 ^ n→i(m) ≠ 0, bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), i % 2 ^ n→i(m) = 0, m = 0, bits = [], lastn(m, int2bits(i)).first

Rule argument

pos: right 6

Additional information

Used simplifier rules: bits2int-first-01: bits ≠ [] ⊦ bits2int(bits) < 0 → bits.first;% ⊦ lastn(n, bits) = [] ↔ n = 0;