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Proof step 2 from proof short-size-01 (previous step)

32768 < 2 ^ (n→i(# int2bits(s2i(sho))) - 1) ⊦ # int2bits(s2i(sho)) = 1

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: n→i(# int2bits(s2i(sho))) - 1 = n→i(# int2bits(s2i(sho)) - 1)

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ n→i(m) = n→i(n) + 1 ↔ m = n + 1;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < 1 ↔ m = 0;% ⊦ ¬ m ≤ n ↔ n < m;%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used formulas: n→i(# int2bits(s2i(sho))) - 1 = n→i(# int2bits(s2i(sho)) - 1)