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Proof step 11 from proof int2int-lastn-negpos (previous step)

# lastn(m, int2bits(i)) ≠ 0 ∧ # bits ≠ 0 ∧ bits.first, bits2int(int2bits(i)) = ~(2 ^ n→i(# lastn(m, int2bits(i))) * bits2uint(bitsinvert(bits))) - bits2uint(bitsinvert(lastn(m, int2bits(i)))) + 1, # bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), 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 12:
# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), ~1 = bits2uint(bitsinvert(bits)) * 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

Rule argument

Additional information

Used simplifier rules: int2int-inj: ⊦ bits2int(int2bits(i)) = i;% ⊦ # bits = 0 ↔ bits = [];% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ ~ i = j ↔ i + j = 0;% ⊦ ~ i - j = ~(i + j);% ⊦ n ≤ # bits → # lastn(n, bits) = n;