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Proof step 36 from proof band-min (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 37:
j = i + 1, # int2bits(i2 / 2) = # int2bits(i / 2), bits2int(bitsand(int2bits(i2 / 2) + (~1 = (i2 + 1) % 2)', int2bits(i / 2) + (~1 = j % 2)')) < i0, i2 < ~1, i2 < 0, j < 0, 2 ^ (n→i(# int2bits(i2 / 2)) + 1) + i0 ≤ 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, j = ~1

Rule argument

seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 1, i] rotate: T spec: int-abs inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i