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Proof step 4 from proof bits2int-nls-01 (previous step)

~(2 ^ n→i(# bits - 1)) ≤ bits2int(bits) ∧ bits2int(bits) < 2 ^ n→i(# bits - 1), # bits ≤ n, 2 ^ n→i(n) + bits2int(bits) < 0 ⊦ bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ 2 ^ n→i(# bits - 1) + bits2int(bits), 2 ^ n→i(n) + bits2int(bits) < 0, # bits ≤ n, bits2int(bits) < 2 ^ n→i(# bits - 1), 2 ^ n→i(n) < 2 ^ n→i(# bits - 1) ⊦ bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j;