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Proof step 33 from proof bits2int-lasteq (previous step)

⊦ bits0 = [], ~1 = 2 * bits2uint(bitsinvert(bits1)) + bits2int(bits0 + true ') ↔ ~1 = bits2int(bits0) + bits2uint(bitsinvert(bits1)), bits0.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] → [bits0, boolvar, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: bits2int-pos: ⊦ bits2int(false ' + bits) = bits2uint(bits);%bits2uint-one: ⊦ bits2uint(bits + true ') = bits2uint(bits) * 2 + 1;% ⊦ i + j = j + i;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 1 = ~1 ↔ i + 1 + 1 = 0;% ⊦ i * j + i = i * (j + 1);% ⊦ 2 * i = 0 ↔ i = 0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits0, boolvar, bits