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Proof step 31 from proof int2int-lastn-negpos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), bits2int(bits) * 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))) = i, bits2int(bits) < 2 ^ n→i(m), bits2int(bits) < 0, m ≤ # int2bits(i), 0 ≤ bits2int(lastn(m, int2bits(i))), 0 < 2 ^ n→i(m), i < 0, i < bits2int(lastn(m, int2bits(i))), i + 1 < 2 ^ n→i(m), i < 2 ^ n→i(m), bits2int(bits) < bits2int(lastn(m, int2bits(i))), bits2int(bits) + 1 < 2 ^ n→i(m), 2 ^ n→i(m) ≤ bits2int(lastn(m, int2bits(i))), 0 < bits2int(lastn(m, int2bits(i))), i + 1 < bits2int(lastn(m, int2bits(i))), bits2int(bits) + 1 < bits2int(lastn(m, int2bits(i))) ⊦ bits2int(lastn(m, int2bits(i))) = bits2int(lastn(m, int2bits(i))) % 2 ^ n→i(m), i % 2 ^ n→i(m) = 0, m = 0, bits = [], bits2int(lastn(m, int2bits(i))) % 2 ^ n→i(m) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0;% ⊦ i +1 = i + 1;% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i * j = j * i;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 ≤ i → abs(i) = i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;