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Proof step 24 from proof bitsor-bi-zeroes (previous step)

bits3.first ⊦ # bits3 = 1, bits2int(bits3 + bits1) = bits2uint(bits1), bits3 = [], true ∈ bits3

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] → [bits3, boolvar, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: ⊦ boolvar + bits = boolvar ' + bits;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ boolvar ' + bits ≠ [];%⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits3, boolvar, bits