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Proof step 41 from proof and-insgn (previous step)

¬(bits2 + boolvar0 ' ≠ [] ∧ boolvar1 ' = true '), ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bitsand(boolvar1 ', bits2 + boolvar0 ') ↔ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits0 + boolvar1 ' = (bits0 + boolvar1 ').first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', (¬ boolvar1) ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
true ∈ bitsand(bits0, bits2), ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ bits0 + false ' = (bits0 + false ').first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first ', true ∈ bits0

Rule argument

Additional information

Used simplifier rules: bitsand-zerol: ⊦ bitsand(false ', bits + boolvar ') = false ';% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];