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Proof step 4 from proof bint-and-011x-bits2unit (previous step)

¬(i ∈bint ∧ 0 ≤ i), ~2147483648 ≤ i, i ≤ 2147483647 ⊦ bits2uint(fillsgn(32, int2bits(i))) = band(4294967295, i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
~2147483648 ≤ i, i < 0 ⊦ bits2uint(fillsgn(32, int2bits(i))) = band(4294967295, i)

Rule argument

Additional information

Used simplifier rules: inbint: ⊦ i ∈bint ↔ ~2147483648 ≤ i ∧ i ≤ 2147483647;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ ¬ i ≤ j ↔ j < i;%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);