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Proof step 10 from proof band-min (previous step)

i = 0, # int2bits(i) = # int2bits(j), i0 + 2 ^ n→i(# int2bits(j)) ≤ 0, j ≤ 0, i ≤ 0, bits2int(bitsand(int2bits(i), int2bits(j))) < i0, ∀ i0, j, i1. i→n(i1) < i→n(i) → # int2bits(i1) = # int2bits(j) ∧ bits2int(bitsand(int2bits(i1), int2bits(j))) < i0 ∧ i1 ≤ 0 ∧ j ≤ 0 ∧ i0 + 2 ^ n→i(# int2bits(j)) ≤ 0 → false ⊦ i ≠ 0 ∧ i ≠ ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ j ≤ i ∧ i < 0 → j < 0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ # boolvar ' = 0 +1;% ⊦ int2bits(0) = false ';