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Proof step 41 from proof sis-on-bits (previous step)

16 ≤ # int2bits(i) ⊦ s2i(i2s(bits2int(lastn(16, int2bits(i))))) = bits2int(lastn(16, int2bits(i))), ~32768 ≤ i ∧ i ≤ 32767

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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