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Proof step 57 from proof and-ass (previous step)

1 < #(bitsand(bits4, bits) + (boolvar0 ∧ boolvar1)') ∧ 1 < #(bits3 + boolvar '), ∀ bits2, bits0, bits1. # bits2 < #(bits4 + boolvar0 ') → true → bits2 = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsand(bits2, bitsand(bits0, bits1)) = bitsand(bitsand(bits2, bits0), bits1) ⊦ bits + boolvar1 ' = (bits + boolvar1 ').first ', bits3 + boolvar ' = (bits3 + boolvar ').first ', bitsand(bits4, bitsand(bits, bits3)) + (boolvar0 ∧ boolvar1 ∧ boolvar)' = bitsand((bitsand(bits4, bits) + (boolvar0 ∧ boolvar1)').butlast, (bits3 + boolvar ').butlast) + ((bitsand(bits4, bits) + (boolvar0 ∧ boolvar1)').last ∧ (bits3 + boolvar ').last), bits4 = [], bits4 + boolvar0 ' = bits4.first ', bitsand(bits, bits3) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
∀ bits2, bits0, bits1. # bits2 < #(bits4 + boolvar0 ') → true → bits2 = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsand(bits2, bitsand(bits0, bits1)) = bitsand(bitsand(bits2, bits0), bits1) ⊦ bitsand(bits4, bitsand(bits, bits3)) + (boolvar0 ∧ boolvar1 ∧ boolvar)' = bitsand(bitsand(bits4, bits), bits3) + ((boolvar0 ∧ boolvar1) ∧ boolvar)', bitsand(bits4, bits) = [], bits3 = [], bits4 = [], bits4 + boolvar0 ' = bits4.first ', bitsand(bits, bits3) = [], bits + boolvar1 ' = (bits + boolvar1 ').first ', bits3 + boolvar ' = bits3.first '

Rule argument

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # 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 + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;