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Proof step 21 from proof intshiftrsign-if (previous step)

¬(sublist(0, n, int2bits(i)) ≠ [] ∧ (sublist(0, n, int2bits(i)).first ↔ ¬ 0 ≤ i)), # int2bits(i) = n + i→n(band(j, 31)), i→n(band(j, 31)) < # int2bits(i) ⊦ bits2int(mklist(¬ 0 ≤ i, i→n(band(j, 31))) + sublist(0, n, int2bits(i))) = bits2int(sublist(0, n, int2bits(i))), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
# int2bits(i) = n + i→n(band(j, 31)), i→n(band(j, 31)) < # int2bits(i) ⊦ bits2int(mklist(¬ 0 ≤ i, i→n(band(j, 31))) + sublist(0, n, int2bits(i))) = bits2int(sublist(0, n, int2bits(i))), n = 0, sublist(0, n, int2bits(i)).first ↔ ¬ 0 ≤ i

Rule argument

Additional information

Used simplifier rules: ⊦ sublist(0, n, bits) = [] ↔ n = 0;