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Proof step 78 from proof band-max (previous step)

1 < #(int2bits(i / 2) + (i % 2 = 1)') ∧ 1 < #(int2bits(j / 2) + (j % 2 = 1)'), 0 < i, 1 < i, 0 ≤ i, i < bits2int ( bitsand((int2bits(i / 2) + (i % 2 = 1)').butlast, (int2bits(j / 2) + (j % 2 = 1)').butlast) + ((int2bits(i / 2) + (i % 2 = 1)').last ∧ (int2bits(j / 2) + (j % 2 = 1)').last)), 1 < j, 0 < j, 1 < bits2int ( bitsand((int2bits(i / 2) + (i % 2 = 1)').butlast, (int2bits(j / 2) + (j % 2 = 1)').butlast) + ((int2bits(i / 2) + (i % 2 = 1)').last ∧ (int2bits(j / 2) + (j % 2 = 1)').last)), 0 < bits2int ( bitsand((int2bits(i / 2) + (i % 2 = 1)').butlast, (int2bits(j / 2) + (j % 2 = 1)').butlast) + ((int2bits(i / 2) + (i % 2 = 1)').last ∧ (int2bits(j / 2) + (j % 2 = 1)').last)), ∀ 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

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + boolvar ').butlast = bits;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;