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Proof step 81 from proof getsetShort-theorem (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 82:
j % 256 = 0, j / 256 ≤ 127, ~128 ≤ j / 256, ~32768 ≤ j, j < 0, ~256 < b2i(i2b(j)), b2i(i2b(j)) < 0 ⊦

Rule argument

Additional information

Used simplifier rules: byte-neq: i < ~128 ⊦ (* i) ≠ b2i(by);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%numls-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%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs