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Proof step 8 from proof bits2int-zeros-01 (previous step)

⊦ bits = [], bits2uint(bits.rest + false ' + bits0) = bits2uint(bits0), true ∈ bits, bits.first

Heuristic var 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, bits1] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: bits2uint-zeros-02: ⊦ ¬ true ∈ bits → bits2uint(bits + bits0) = bits2uint(bits0);%bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar, bits1