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Proof step 62 from proof getsetShort-theorem (previous step)

j / 256 = i + 1, ub2i(i2b(j)) < 256, ~32768 ≤ j, j < 0, 0 ≤ ub2i(i2b(j)), j / 256 ≤ 128, ~127 ≤ j / 256 ⊦ ub2i(i2b(j)) + 256 * i = j, j % 256 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 63:
j / 256 = i + 1, 0 ≤ ub2i(i2b(j)), j < 0, ~32768 ≤ j, ub2i(i2b(j)) < 256, ~128 ≤ i, i ≤ 127 ⊦ j % 256 = 0, ub2i(i2b(j)) + 256 * i = j

Rule argument

Additional information

Used simplifier rules: numle-03: ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs