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Proof step 65 from proof getsetShort-theorem (previous step)

¬ b2i(i2b(j)) < 0, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
j / 256 = i + 1, 0 ≤ b2i(i2b(j)), b2i(i2b(j)) < 256, ~32768 ≤ j, j < 0, i ≤ 127, ~128 ≤ i ⊦ 256 * i + b2i(i2b(j)) = j, j % 256 = 0

Rule argument

Additional information

Used simplifier rules: ubyte-pos: ⊦ 0 ≤ b2i(by) → ub2i(by) = b2i(by);%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; 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