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Proof step 4 from proof byte-size (previous step)

0 < 128 ∧ 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

Proof step 5:
# int2bits(128) ≤ i→n(n→i(# int2bits(b2i(by))) - 1) + 1 ⊦ # int2bits(b2i(by)) = 1

Rule argument

Additional information

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