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Proof step 4 from proof int2bits-rec-neg-03 (previous step)

j < 0 ∧ j ≠ ~1, j < ~1, j < 0 ⊦ int2bits((j - 1) / 2) + (~1 = j % 2)' = int2bits(j / 2 - 1) + true ', j % 2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j < ~1, j < 0 ⊦ j % 2 = 0, j = ~1, j / 2 = (j - 1) / 2 + 1

Rule argument

Additional information

Used simplifier rules: int2bits-inj: ⊦ int2bits(i) = int2bits(j) ↔ i = j;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;% ⊦ j < 0 → (j % 2 ≠ ~1 ↔ j % 2 = 0);