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Proof step 27 from proof int2int-lastn-negneg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
i = (bits2int(bits) + 1) * 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))), int2bits(i) = bits + lastn(m, int2bits(i)), # bits = # int2bits(i) - m, bits2int(lastn(m, int2bits(i))) + 1 < 2 ^ n→i(m), i < 2 ^ n→i(m), i + 1 < 2 ^ n→i(m), 0 < 2 ^ n→i(m), bits2int(lastn(m, int2bits(i))) < 0, i < 0, m ≤ # int2bits(i), bits.first, bits2int(lastn(m, int2bits(i))) < 2 ^ n→i(m) ⊦ bits = [], m = 0, bits2int(lastn(m, int2bits(i))) = bits2int(lastn(m, int2bits(i))) % 2 ^ n→i(m), i + 2 ^ n→i(m) = 0

Rule argument

Additional information

Used simplifier rules: bits2int-first-01: bits ≠ [] ⊦ bits2int(bits) < 0 → bits.first;% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i +1 = i + 1;% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ lastn(n, bits) = [] ↔ n = 0;