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Proof step 36 from proof sis-on-bits (previous step)

0 < i, i < 2 ^ (n→i(16) - 2) ⊦ 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: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)