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Proof step 15 from proof sis-on-bits (previous step)

¬ i < ~1, 16 < # int2bits(i), ~32768 ≤ i, i ≤ 0 ⊦ i = bits2int(lastn(16, int2bits(i))), 16 = # fillsgn(16, int2bits(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
16 < # int2bits(i), i ≤ 0, ~1 ≤ i ⊦ i = bits2int(lastn(16, int2bits(i))), 16 = # fillsgn(16, int2bits(i))

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;