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Proof step 14 from proof int2int-lastn-negpos (previous step)

bits ≠ [] ∧ bits.first, # bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), ~1 = (~ bits2int(bits) - 1) * 2 ^ n→i(m) + bits2uint(bitsinvert(lastn(m, int2bits(i)))) + i, bits.first, i < 0, m ≤ # int2bits(i) ⊦ bits = [], bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), i % 2 ^ n→i(m) = 0, m = 0, lastn(m, int2bits(i)).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), (bits2int(bits) + 1) * 2 ^ n→i(m) = (i + bits2uint(bitsinvert(lastn(m, int2bits(i))))) + 1, 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

Rule argument

Additional information

Used simplifier rules: ⊦ i + ~1 = i - 1;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + ~ j = i - j;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ~ i * j = ~(i * j);% ⊦ ~ i - j = ~(i + j);