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Proof step 18 from proof bint-and-011x-bits2unit (previous step)

i ∈bint, ~2147483648 ≤ i, i < 0 ⊦ bits2int(fillsgn(32, int2bits(i))) + 2 ^ n→i(32) = 4294967296 + i, fillsgn(32, int2bits(i)) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
~2147483648 ≤ i, i < 0 ⊦ fillsgn(32, int2bits(i)) = [], bits2int(fillsgn(32, int2bits(i))) = i

Rule argument

Additional information

Used simplifier rules: inbint: ⊦ i ∈bint ↔ ~2147483648 ≤ i ∧ i ≤ 2147483647;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = i + i0 ↔ j = i0;