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Proof step 7 from proof byte-size (previous step)

0 < n→i(# int2bits(b2i(by))) - 1, # int2bits(128) ≤ i→n((n→i(# int2bits(b2i(by))) - 1) + 1) ⊦ # int2bits(b2i(by)) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i→n(n→i(n)) = n;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i) = j;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ i < j - i0 ↔ i + i0 < j;