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Proof step 7 from proof uint2bits-expos0 (previous step)

uint2bits(i) = boolvar ' + bits0, 0 < i, boolvar ∈ uint2bits(i) ⊦ uint2bits(i) = false ' + true ' + bits0.rest, uint2bits(i).first, ∃ bits. uint2bits(i) = false ' + true ' + bits

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [uint2bits(i), boolvar0, bits1] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: uint2bits(i), boolvar0, bits1