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Proof step 3 from proof uint2bits-falsein-01 (previous step)

0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ false ∈ uint2bits(j).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

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 < j ∧ j ≠ 1, 0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ false ∈ (uint2bits(j / 2) + (j % 2 = 1)).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Rule argument

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

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ (boolvar ' + bits).rest = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ 2 ^ i = 2 ↔ i = 1; Used terms: j