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Proof step 56 from proof and-sgn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 57:
⊦ ¬ false ∈ bits ∧ bits ≠ [] ∧ bits0 ≠ [] → (boolvar ∈ bitsand(bits, bits0) ↔ boolvar ∈ bits0)
Proof step 58:
¬ false ∈ bits0 ∧ bits0 ≠ [] ∧ bits2 ≠ [], ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits0 = [], bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 + true ' = bits0.first ', sgn(bits2 + boolvar ') = boolvar ', (¬ boolvar) ∈ bits2, false ∈ bits0

Rule argument

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

Additional information

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