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Proof step 3 from proof sis-on-bits (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
~32768 ≤ i, i ≤ 32767 ⊦ i = bits2int(lastn(16, fillsgn(16, int2bits(i))))

Rule argument

Additional information

Used simplifier rules: sis-in: ⊦ i ∈short → s2i(i2s(i)) = i;%inshort: ⊦ i ∈short ↔ ~32768 ≤ i ∧ i ≤ 32767;% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)