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Proof step 3 from proof uint2bits-second (previous step)

0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → uint2bits(i0).rest.first ⊦ uint2bits(i).rest.first

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i ∧ i ≠ 1 → uint2bits(i) = uint2bits(i / 2) + (i % 2 = 1)
Proof step 5:
0 < i ∧ i ≠ 1, 0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 < i0 → uint2bits(i0).rest.first ⊦ (uint2bits(i / 2) + (i % 2 = 1)).rest.first

Rule argument

name: uint2bits-rec-01 rotate: F subst: [i] → [i]

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ (boolvar ' + bits).rest = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar '.first ↔ boolvar; Used terms: i