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Proof step 22 from proof and-insgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
∀ bits1, bits, boolvar. # bits1 < # false ' → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ bits = [], boolvar ∈ bitsand(sgn(bits), false ') ↔ ¬ boolvar

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ' ≠ [];% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;