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Proof step 72 from proof byte-mult-256 (previous step)

256 * (i0 + i) + j0 = (i0 + i + 1) * 256 + (j0 - 256), i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, i2b(256 * (i0 + i) + j0) = i2b(j0)

User applied rule insert equation

And obtained the following new goals

Proof step 73:
i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, i2b((i0 + i + 1) * 256 + (j0 - 256)) = i2b(j0)

Rule argument

pos: left 1 flags: discard

Proof step 72 from proof byte-mult-256 (previous step)

User applied rule insert equation

And obtained the following new goals

Rule argument

Additional information