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Proof step 22 from proof int2bytes-rec-neg-03 (previous step)

0 < i, 2 ^ (i * 8) * (j0 + 1) < 0, j0 < 0 ⊦ 2 ^ (i * 8) = 0, int2bytes(j0) = int2bytes(j0 + (1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
abs(i) < abs(j) ⊦ i / j = 0
Proof step 24:
¬ abs(1 - 2 ^ (i * 8)) < abs(2 ^ (i * 8)), 0 < i, 2 ^ (i * 8) * (j0 + 1) < 0, j0 < 0 ⊦ 2 ^ (i * 8) = 0, int2bytes(j0) = int2bytes(j0 + (1 - 2 ^ (i * 8)) / 2 ^ (i * 8))

Rule argument

seq: abs(i) < abs(j) ⊦ i / j = 0 subst: [j, i] → [2 ^ (i * 8), 1 - 2 ^ (i * 8)] rotate: F spec: int-div inst: name: div-zero-02

Additional information

Used simplifier rules: ⊦ i + 0 = i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: 2 ^ (i * 8), 1 - 2 ^ (i * 8)