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Proof step 52 from proof bib-neg-high (previous step)

i0 < 0 ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i0 * 256))))) = 0, ~256 ≠ i0 * 256 ∧ lastn(8, int2bits(i0 * 256)).first

User applied rule cut formula

And obtained the following new goals

Proof step 53:
256 = 2 ^ 8, i0 < 0 ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i0 * 256))))) = 0, ~256 ≠ i0 * 256 ∧ lastn(8, int2bits(i0 * 256)).first

Rule argument

fma: 256 = 2 ^ 8

Additional information

Used formulas: 256 = 2 ^ 8