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Proof step 78 from proof getsetShort-theorem (previous step)

¬ b2i(i2b(j)) < 0, j % 256 = 0, j / 256 ≤ 127, ~128 ≤ j / 256, ub2i(i2b(j)) < 256, ~32768 ≤ j, j < 0, 0 < ub2i(i2b(j)) ⊦ ub2i(i2b(j)) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 79:
j % 256 = 0, j / 256 ≤ 127, ~128 ≤ j / 256, b2i(i2b(j)) < 256, ~32768 ≤ j, j < 0, 0 < b2i(i2b(j)) ⊦ b2i(i2b(j)) = 0

Rule argument

Additional information

Used simplifier rules: ubyte-pos: ⊦ 0 ≤ b2i(by) → ub2i(by) = b2i(by);%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs