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Proof step 25 from proof and-true (previous step)

bitsand(bits.butlast, boolvar ' + bits1.butlast) = [], bits.last, bits1.last ⊦ bits = bits.first ', bits = [], bits = true ', bits1 = []

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, boolvar0, bits0] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: and-notempty-05: ⊦ bitsand(boolvar0 ' + bits, boolvar ' + bits0) ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bits = bits + bits0 ↔ bits0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → (bits0 + bits).butlast = bits0 + bits.butlast;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar0, bits0