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Proof step 19 from proof byte-mult-256 (previous step)

256 * i + j0 = (i - 1) * 256 + 256 + j0, j0 < 0, ~256 < j0, 0 < i ⊦ i2b(256 * i + j0) = i2b(j0), i = 0

User applied rule insert equation

And obtained the following new goals

Proof step 20:
j0 < 0, ~256 < j0, 0 < i ⊦ i2b((i - 1) * 256 + 256 + j0) = i2b(j0), i = 0

Rule argument

pos: left 1 flags: discard

Additional information