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Proof step 68 from proof getsetShort-theorem (previous step)

¬(j < 0 ∧ ~128 ≤ j % 256), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 69:
j / 256 = i + 1, ~32768 ≤ j, j < 0, i ≤ 127, ~128 ≤ i, j % 256 < ~128, ~256 ≤ j % 256, j % 256 < i ⊦ j % 256 = 0, 256 * (i + 1) + j % 256 = j

Rule argument

Additional information

Used simplifier rules: bib-neg-low: ⊦ i < 0 ∧ i % 256 < ~128 → b2i(i2b(i)) = i % 256 + 256;%numlslsdel: j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0); from specification int-basic1%numlsledel-01: i0 ≠ 0, i0 < i ⊦ (* i) ≤ j → (* i0) < j; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%mod-ls-zero: 0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0; from specification int-div%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%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%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); 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