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Proof step 25 from proof sis-on-bits (previous step)

¬ 16 ≤ # int2bits(i) ⊦ s2i(i2s(bits2int(lastn(16, int2bits(i))))) = bits2int(lastn(16, fillsgn(16, int2bits(i)))), i ∈short

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;