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Proof step 60 from proof bits2int-pos (previous step)

~128 ≤ bits2uint(firstn(# bits0 % 8, bits0)) ∧ bits2uint(firstn(# bits0 % 8, bits0)) ≤ 127, 7 < # bits0, # bits0 % 8 ≤ 7, # bits0 % 8 < # bits0, bits2int(fillsgn(8, int2bits(bits2uint(firstn(# bits0 % 8, bits0)))) + bytes2bits(bits2bytes(restn(# bits0 % 8, bits0)))) < 0 ⊦ bits0 = [], (# bits0 + 1) % 8 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)