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

32768 < 2 ^ n→i(# int2bits(s2i(sho)) - 1) ⊦ # int2bits(s2i(sho)) - 1 = 0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
32768 = 2 ^ 15, 32768 < 2 ^ n→i(# int2bits(s2i(sho)) - 1) ⊦ # int2bits(s2i(sho)) - 1 = 0

Rule argument

fma: 32768 = 2 ^ 15

Additional information

Used formulas: 32768 = 2 ^ 15