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Proof step 12 from proof bitsinvert-size (previous step)

bitsinvert(bits0) = boolvar0 ' + bits, # sgn(bitsinvert(bits0)) = # sgn(bits0), (¬ boolvar0) ∈ bits0 ⊦ boolvar0 ↔ ¬ boolvar, bits0 = [], # sgn((¬ boolvar)' + bitsinvert(bits0)) = # sgn(boolvar ' + bits0), bitsinvert(bits0).first ↔ boolvar

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] → [bitsinvert(bits0), boolvar1, bits1] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: inv-empty: ⊦ bitsinvert(bits) = [] ↔ bits = [];%inv-empty: ⊦ bitsinvert(bits) = [] ↔ bits = [];% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ bits = boolvar ' + bits0 ∧ bits = boolvar0 ' + bits1 → (boolvar ↔ boolvar0) ∧ bits0 = bits1; (forward)%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bitsinvert(bits0), boolvar1, bits1