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Proof step 5 from proof bits2int-rec-false (previous step)

bits.first ⊦ bits2int(bits + false ') = bits2int(bits) * 2, bits = []

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

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits2int(true ' + bits) = ~ bits2uint(bitsinvert(true + bits) +1);%bits2uint-zero: ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ i * j = j * i;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ bitsinvert(false ') = true ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + true ')+1 = bits +1 + false;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ ~ i * j = ~(i * j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ~ i = ~ j ↔ i = j;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar, bits0