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Proof step 19 from proof bitsinvert-size (previous step)

bitsinvert(bits0) = boolvar ' + bits1, # sgn(bitsinvert(bits0)) = # sgn(bits0), (¬ boolvar) ∈ bits0 ⊦ bits0 = [], 1 + # bits0 = # sgn(boolvar ' + bits0), 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] → [bits0, boolvar0, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: inv-append: ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);%inv-one: ⊦ bitsinvert(boolvar ') = (¬ boolvar)';% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ boolvar ' + bits = boolvar0 ' + bits0 ↔ (boolvar ↔ boolvar0) ∧ bits = bits0;% ⊦ boolvar ' + bits ≠ [];% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits0, boolvar0, bits