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Proof step 61 from proof byte-mult-256 (previous step)

256 ≠ 0, ∃ j0. j = j0 * 256, j < 0, 0 < i ⊦ i = 0, i2b(i * 256 + j) = i2b(j)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: byte-mult: ⊦ i2b(i * 256) = i2b(0);%num-i2b-equal: ~128 ≤ i, i ≤ 127, ~128 ≤ j, j ≤ 127 ⊦ i2b((* i)) = i2b((* j)) ↔ i = j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i * j0 = i * (j + j0);