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Proof step 15 from proof bitsand-app (previous step)

1 < #(bits1 + boolvar0 ') ∧ 1 < #(bits4 + boolvar '), # bits1 = # bits4, ∀ bits2, bits0, bits, bits3. # bits3 < #(bits4 + boolvar ') → # bits = # bits3 → bits0 = [] ∨ bits2 = [] ∨ bits = [] ∨ bitsand(bits0 + bits, bits2 + bits3) = bitsand(bits0, bits2) + bitsand(bits, bits3) ⊦ bits2 = [], bits0 = [], bitsand(bits0 + bits1, bits2 + bits4) + (boolvar0 ∧ boolvar)' = bitsand(bits0, bits2) + (bitsand((bits1 + boolvar0 ').butlast, (bits4 + boolvar ').butlast) + ((bits1 + boolvar0 ').last ∧ (bits4 + boolvar ').last))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# bits1 = # bits4, ∀ bits2, bits0, bits, bits3. # bits3 < #(bits4 + boolvar ') → # bits = # bits3 → bits0 = [] ∨ bits2 = [] ∨ bits = [] ∨ bitsand(bits0 + bits, bits2 + bits3) = bitsand(bits0, bits2) + bitsand(bits, bits3) ⊦ bits2 = [], bits0 = [], bitsand(bits0 + bits1, bits2 + bits4) = bitsand(bits0, bits2) + bitsand(bits1, bits4), bits1 = [], bits4 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # 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);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;