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Proof step 2 from proof inv-false-01 (previous step)

bitsinvert(bits) = false ' ⊦ bits = true '

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: bits

Additional information

Used simplifier rules: invert-base: ⊦ bitsinvert([]) = [];%inv-append: ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);%inv-one: ⊦ bitsinvert(boolvar ') = (¬ boolvar)';%inv-empty: ⊦ bitsinvert(bits) = [] ↔ bits = [];%bitlist freely generated by [], +;;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ boolvar ' ≠ [];