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Proof step 62 from proof byte-mult-256 (previous step)

⊦ j < 0, b2i(i2b(i * 256 + j)) = b2i(i2b(j)), i = 0, ¬ i < 0 ∧ ¬ j < 0

User applied rule cut formula

And obtained the following new goals

Proof step 63:
⊦ j % 256 < 0, j < 0, b2i(i2b(i * 256 + j)) = b2i(i2b(j)), i = 0, ¬ i < 0 ∧ ¬ j < 0

Rule argument

fma: j % 256 < 0

Additional information

Used simplifier rules: byte-equal: ⊦ b2i(byi) = b2i(byj) ↔ byi = byj;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 0 < j ∧ 0 ≤ i → ¬ i % j < 0; Used formulas: j % 256 < 0