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Proof step 3 from proof int2bits-mod2-neg (previous step)

i = i0 * 2 + j ∧ j ≤ 0 ∧ abs(j) < 2 ∧ i0 ≤ 0 ∧ 0 < 2 ∧ i < 0, i ≤ 0 ⊦ int2bits(j) = int2bits(i).last '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
abs(j) < 2, j ≤ 0, i0 ≤ 0, i0 * 2 + j < 0 ⊦ int2bits(j) = int2bits(i0 * 2 + j).last '

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;