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Proof step 23 from proof bits2int-lasteq (previous step)

bits0.first ⊦ bits0 = [], bits2uint(bits1) * 2 = bits2int(bits0 + false ') ↔ bits2uint(bits1) = bits2int(bits0)

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] → [bits0, boolvar, bits] 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;% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ bitsinvert(false ') = true ';% ⊦ (bits + true ')+1 = bits +1 + false;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ ~ i = j ↔ i + j = 0;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 2 * i = 0 ↔ i = 0;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits0, boolvar, bits