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Proof step 63 from proof bits2int-pos (previous step)

0 ≤ bits2uint(firstn(# bits0 % 8, bits0)), ~128 ≤ bits2uint(firstn(# bits0 % 8, bits0)), bits2uint(firstn(# bits0 % 8, bits0)) ≤ 127, 7 < # bits0, # bits0 % 8 ≤ 7, # bits0 % 8 < # bits0, bits2int(fillsgn(8, uint2bits(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: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;