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Proof step 1 from proof bor-256-plus-01 (conclusion)

⊦ j0 < 256 ∧ 0 ≤ j0 → bor(j * 256, j0) = j * 256 + j0

User applied rule cut formula

And obtained the following new goals

Proof step 2:
256 = 2 ^ 8 ⊦ j0 < 256 ∧ 0 ≤ j0 → bor(j * 256, j0) = j * 256 + j0

Rule argument

fma: 256 = 2 ^ 8

Additional information

Used formulas: 256 = 2 ^ 8