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Proof step 75 from proof byte-mult-256 (previous step)

¬(i0 + i + 1 < 0 ∧ j0 - 256 < 0), i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, i2b((i0 + i + 1) * 256 + (j0 - 256)) = i2b(j0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 76:
i0 + i = ~1, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i = 0, i2b((i0 + i + 1) * 256 + j0 - 256) = i2b(j0)

Rule argument

Additional information

Used simplifier rules: i2b-noteq: ⊦ i2b(i) ≠ i2b(j) → i ≠ j;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i * j = i ↔ i = 0 ∨ j = 1;% ⊦ i + j = i + i0 ↔ j = i0;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i0) = i + j - i0;