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Proof step 12 from proof uint2bits-second (previous step)

uint2bits(i / 2).rest = [], i % 2 = 0, 0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → uint2bits(i0).rest.first ⊦ i = 1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
uint2bits(i / 2) = boolvar + bits ∧ uint2bits(i / 2) ≠ [], bits = [], i % 2 = 0, 0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → uint2bits(i0).rest.first ⊦ i = 1

Rule argument

seq: ⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0) subst: [bits, boolvar, bits0] → [uint2bits(i / 2), boolvar, bits] rotate: T spec: list-data inst: bitlista name: elim-1

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;%⊦ bits ≠ [] → ( (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0); (elim) Used terms: uint2bits(i / 2), boolvar, bits