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Proof step 40 from proof band-min (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ 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;