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Proof step 30 from proof int2int-lastn-negneg (previous step)

¬ abs(bits2int(lastn(m, int2bits(i)))) < 2 ^ n→i(m), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), i = (bits2int(bits) + 1) * 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))), 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))) ≤ 0, bits2int(lastn(m, int2bits(i))) < 2 ^ n→i(m), bits.first, m ≤ # int2bits(i), i < 0, 0 < 2 ^ n→i(m), i + 1 < 2 ^ n→i(m), i < 2 ^ n→i(m), bits2int(lastn(m, int2bits(i))) + 1 < 2 ^ n→i(m), bits2int(lastn(m, int2bits(i))) < 0 ⊦ i + 2 ^ n→i(m) = 0, bits2int(lastn(m, int2bits(i))) = bits2int(lastn(m, int2bits(i))) % 2 ^ n→i(m), m = 0, bits = []

Rule argument

Additional information

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