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Proof step 59 from proof byte-mult-256 (previous step)

j % 256 = 0, j < 0, 0 < i ⊦ i = 0, i2b(i * 256 + j) = i2b(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 60:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 61:
256 ≠ 0, ∃ j0. j = j0 * 256, j < 0, 0 < i ⊦ i = 0, i2b(i * 256 + j) = i2b(j)

Rule argument

seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [256, j] rotate: F spec: int-div inst: name: modex

Additional information

Used terms: 256, j