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Proof step 6 from proof short-size (previous step)

0 ≤ 15, 15 < n→i(# int2bits(s2i(sho)) - 1) ⊦ # int2bits(s2i(sho)) - 1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)%n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ int2bits(j) ≠ [];%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;