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Proof step 10 from proof shiftright-minus-one (previous step)

⊦ bits2int((mklist(true, i→n(j)) + true ').first ') = ~1

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] → [mklist(true, i→n(j)) + true ', boolvar, bits] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bits + boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ ¬ false ∈ mklist(true, n);% ⊦ bits2int(true ') = ~1;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: mklist(true, i→n(j)) + true ', boolvar, bits