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Proof step 12 from proof sis-on-bits (previous step)

¬ 0 < i, 16 < # int2bits(i), ~32768 ≤ i, i ≤ 32767 ⊦ i = bits2int(lastn(16, int2bits(i))), 16 = # fillsgn(16, int2bits(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
16 < # int2bits(i), ~32768 ≤ i, i ≤ 0 ⊦ i = bits2int(lastn(16, int2bits(i))), 16 = # fillsgn(16, int2bits(i))

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);% ⊦ ¬ i < j ↔ j ≤ i;