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Proof step 36 from proof and-ass (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
bits.first, ∀ bits2, bits0, bits1. # bits2 < # bits.first ' → true → bits2 = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsand(bits2, bitsand(bits0, bits1)) = bitsand(bitsand(bits2, bits0), bits1) ⊦ bits3 + boolvar ' = (bits3 + boolvar ').first ', bits4 + boolvar0 ' = (bits4 + boolvar0 ').first ', bitsand(bits3, bits4) + (boolvar ∧ boolvar0)' = bitsand(bits3 + boolvar ', bits4 + boolvar0 ')

Rule argument

Additional information

Used simplifier rules: bitsand-onel: ⊦ bitsand(true ', bits + boolvar ') = bits + boolvar;% ⊦ bits + boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;