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Proof step 46 from proof getsetShort-theorem (previous step)

j = i * 256 + j0 ∧ 0 ≤ j0 ∧ j0 < 256 ∧ 0 ≤ i ∧ 0 < 256, 0 < i, 127 < i, j < 32768, 0 ≤ j, 0 ≤ ub2i(i2b(j)), ub2i(i2b(j)) < 256 ⊦ 256 * b2i(i2b(i)) + ub2i(i2b(j)) = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: byte-mult-256: ⊦ i2b(i * 256 + j) = i2b(j);%ubyte-pos-03: ⊦ 0 ≤ i ∧ i < 256 → ub2i(i2b(i)) = i;%numlslsdel-01: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; from specification int-basic1%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%numlelsdel: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-ge-add: ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); from specification int-mult%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%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