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Proof step 10 from proof intshiftright-ls-01 (previous step)

abs(j) < 32, j < 32, 0 ≤ j ⊦ bits2int(sublist(0, # int2bits(i), mklist(i < 0, i→n(j)) + int2bits(i))) < 0 ↔ ¬ 0 ≤ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
j < 32, 0 ≤ j ⊦ bits2int(sublist(0, # int2bits(i), mklist(i < 0, i→n(j)) + int2bits(i))) < 0 ↔ ¬ 0 ≤ i

Rule argument

Additional information

Used simplifier rules: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);% ⊦ 0 ≤ i → abs(i) = i;