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Proof step 79 from proof byte-mult-256 (previous step)

j0 = i1 + 256, i0 + i = ~1, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i2b(i1) = i2b(j0), i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 80:
i0 + i = ~1, i1 < 0, ~256 ≤ i1, i < 0, 0 ≤ i0 ⊦ i = 0, i2b(i1) = i2b(i1 + 256)

Rule argument

Additional information

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