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Proof step 81 from proof band-max (previous step)

¬(¬(i % 2 = 1 ∧ j % 2 = 1) ∧ bitsand(int2bits(i / 2), int2bits(j / 2)) ≠ []), 0 < i, 1 < i, 0 ≤ i, i < bits2int(bitsand(int2bits(i / 2), int2bits(j / 2)) + (i % 2 = 1 ∧ j % 2 = 1)'), 1 < j, 0 < j, 1 < bits2int(bitsand(int2bits(i / 2), int2bits(j / 2)) + (i % 2 = 1 ∧ j % 2 = 1)'), 0 < bits2int(bitsand(int2bits(i / 2), int2bits(j / 2)) + (i % 2 = 1 ∧ j % 2 = 1)'), ∀ i0, j. i→n(i0) < i→n(i) → i0 < bits2int(bitsand(int2bits(i0), int2bits(j))) ∧ 0 ≤ i0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 82:
i % 2 = 1, j % 2 = 1, 0 < i, 1 < i, 0 ≤ i, i < bits2int(bitsand(int2bits(i / 2), int2bits(j / 2)) + true '), 1 < j, 0 < j, 1 < bits2int(bitsand(int2bits(i / 2), int2bits(j / 2)) + true '), 0 < bits2int(bitsand(int2bits(i / 2), int2bits(j / 2)) + true '), ∀ i0, j. i→n(i0) < i→n(i) → i0 < bits2int(bitsand(int2bits(i0), int2bits(j))) ∧ 0 ≤ i0 → false ⊦

Rule argument

Additional information

Used simplifier rules: and-notempty-09: ⊦ bitsand(int2bits(i), int2bits(i0)) ≠ [];% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)