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Proof step 6 from proof uint2bits-second (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ bits ≠ [] → (bits + bits0).rest = bits.rest + bits0
Proof step 8:
uint2bits(i / 2) ≠ [], 0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → uint2bits(i0).rest.first ⊦ i = 1, (uint2bits(i / 2).rest + (i % 2 = 1)').first

Rule argument

seq: ⊦ bits ≠ [] → (bits + bits0).rest = bits.rest + bits0 subst: [bits0, bits] → [(i % 2 = 1)', uint2bits(i / 2)] rotate: F spec: list inst: bitlista name: rest-01

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; Used terms: (i % 2 = 1)', uint2bits(i / 2)