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Proof step 2 from proof num-i2b-neg (previous step)

i < ~128 ⊦ i2b(i) = i2b(bits2int(lastn(8, int2bits(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i < ~128, i < 0 ⊦ i2b(i) = i2b(bits2int(lastn(8, int2bits(i))))

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)