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Proof step 29 from proof band-min (previous step)

i = i2 + 1, # int2bits(i2 / 2) + 1 = # int2bits(j), i0 + 2 ^ n→i(# int2bits(j)) ≤ 0, j ≤ 0, bits2int(bitsand(int2bits(i2 / 2) + (~1 = i % 2)', int2bits(j))) < i0, i < 0, ∀ 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 = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
# int2bits(i2 / 2) + 1 = # int2bits(j), i0 + 2 ^ n→i(# int2bits(j)) ≤ 0, j ≤ 0, bits2int(bitsand(int2bits(i2 / 2) + (~1 = (i2 + 1) % 2)', int2bits(j))) < i0, i2 < ~1, i2 < 0, ∀ i0, j, i1. i→n(i1) < i→n(i2 + 1) → # int2bits(i1) = # int2bits(j) ∧ bits2int(bitsand(int2bits(i1), int2bits(j))) < i0 ∧ i1 ≤ 0 ∧ j ≤ 0 ∧ i0 + 2 ^ n→i(# int2bits(j)) ≤ 0 → false ⊦ ~2 = i2

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i + 1 = ~1 ↔ i + 1 + 1 = 0;