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Proof step 21 from proof or-invandinv (previous step)

bits = bits.first ', ∀ bits1, bits0. # bits1 < # bits → true → bits1 = [] ∨ bits0 = [] ∨ bitsor(bits1, bits0) = bitsinvert(bitsand(bitsinvert(bits1), bitsinvert(bits0))) ⊦ bits = false ', bits = true ', bits0 = false ', bits0 = true ', bits = [], bits0 = [], bitsor(bits, bits0) = bitsinvert(bitsand(bitsinvert(bits), bitsinvert(bits0)))

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits, boolvar, bits2] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

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