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Proof step 17 from proof int2int-lastn-negpos (previous step)

lastn(m, int2bits(i)) ≠ [] ∧ ¬ lastn(m, int2bits(i)).first, # bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), (bits2int(bits) + 1) * 2 ^ n→i(m) = (i + (2 ^ n→i(# lastn(m, int2bits(i))) + ~ bits2int(lastn(m, int2bits(i))) - 1)) + 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
2 ^ n→i(m) * bits2int(bits) + 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))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), lastn(m, int2bits(i)).first

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ lastn(n, bits) = [] ↔ n = 0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * (i0 + 1) + j = i + j0 ↔ i * i0 + j = j0;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j - i0 + j = i - i0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i - j - i0 = i - j + i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;% ⊦ n ≤ # bits → # lastn(n, bits) = n;