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Proof step 23 from proof int2int-lastn-negpos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
# 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

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ 0 ≤ n→i(n);% ⊦ 0 < i → (0 < i ^ j ↔ i = 1 ∨ 0 ≤ j);% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ ¬ i < j ↔ j ≤ i;