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Proof step 24 from proof intshiftrsign32 (previous step)

¬ 0 + n0 ≤ # mklist(¬ 0 ≤ i, n1), 32 = n1 + # int2bits(i), 32 = n0 + i→n(band(j, 31)), # int2bits(i) ≤ 32, band(j, 31) ≤ 32 ⊦ band(j, 31) = 0, bits2int(sublist(0, # int2bits(i), mklist(¬ 0 ≤ i, i→n(band(j, 31))) + int2bits(i))) = bits2int((¬ 0 ≤ i)' + sublist(0, n0, mklist(¬ 0 ≤ i, n1) + int2bits(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
32 = n0 + i→n(band(j, 31)), 32 = n1 + # int2bits(i), n1 < n0, band(j, 31) ≤ 32, # int2bits(i) ≤ 32 ⊦ bits2int(sublist(0, # int2bits(i), mklist(¬ 0 ≤ i, i→n(band(j, 31))) + int2bits(i))) = bits2int((¬ 0 ≤ i)' + sublist(0, n0, mklist(¬ 0 ≤ i, n1) + int2bits(i))), band(j, 31) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%mk-len: ⊦ # mklist(boolvar, n) = n; from specification list-last[bitlista]%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ax-08: ⊦ n + 0 = n; from specification nat-basic2