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Proof step 31 from proof bitsand-first-samesize (previous step)

1 < #(boolvar ' + bits2 + boolvar1 ') ∧ 1 < #(boolvar0 ' + bits3 + boolvar2 '), # bits2 = # bits3, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < #(bits3 + boolvar2 ') → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bitsand((boolvar ' + bits2 + boolvar1 ').butlast, (boolvar0 ' + bits3 + boolvar2 ').butlast) + ((boolvar ' + bits2 + boolvar1 ').last ∧ (boolvar0 ' + bits3 + boolvar2 ').last) = (boolvar ∧ boolvar0)' + bitsand(bits2, bits3) + (boolvar1 ∧ boolvar2)', bits3 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (boolvar ' + bits + boolvar0 ').last ↔ boolvar0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;