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Proof step 57 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(b2i(i2b(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: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ bits2uint(bits) < 0;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;