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Proof step 50 from proof and-insgn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 51:
⊦ ¬ false ∈ bits ∧ bits ≠ [] ∧ bits0 ≠ [] → (boolvar ∈ bitsand(bits, bits0) ↔ boolvar ∈ bits0)

Rule argument

name: and-onesin rotate: F subst: [bits0, bits, boolvar] → [bits2, bits0, boolvar]

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;% ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: bits2, bits0, boolvar