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Proof step 70 from proof getsetShort-theorem (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 71:
j / 256 = i + 1, j < 0, ~32768 ≤ j, ~128 ≤ i, i ≤ 127, ~256 ≤ b2i(i2b(j)), b2i(i2b(j)) < 0 ⊦ j % 256 = 0, 256 * (i + 1) + b2i(i2b(j)) = j

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) 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%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; 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%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult