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Proof step 49 from proof and-ass (previous step)

bitsand(bits, bits3) = [], ∀ bits2, bits0, bits1. # bits2 < #(bits4 + boolvar0 ') → true → bits2 = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsand(bits2, bitsand(bits0, bits1)) = bitsand(bitsand(bits2, bits0), bits1) ⊦ bits3 + boolvar ' = (bits3 + boolvar ').first ', bits + boolvar1 ' = (bits + boolvar1 ').first ', bits4 = [], bits4 + boolvar0 ' = bits4.first ', bitsand(bits4 + boolvar0 ', (boolvar1 ∧ boolvar)') = bitsand(bitsand(bits4, bits) + (boolvar0 ∧ boolvar1)', bits3 + boolvar ')

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 50:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ []

Rule argument

name: and-notempty rotate: F subst: [bits0, bits] → [bits3, bits]

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: bits3, bits