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Proof step 6 from proof short-mult-max (previous step)

65536 = 256 * 256, 0 < i, 65536 ≤ i * j, i < 256, j ≤ 256, 65536 < 256 * j, 0 < j ⊦ i = 0

User applied rule insert equation

And obtained the following new goals

Proof step 7:
0 < i, 256 * 256 ≤ i * j, i < 256, j ≤ 256, 256 * 256 < 256 * j, 0 < j ⊦ i = 0

Rule argument

pos: left 1 flags: discard

Additional information