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Proof step 3 from proof uint2bits-div2pot (previous step)

2 ^ i ≤ j, 0 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 → uint2bits(j0 / 2 ^ i) = butlastn(i→n(i), uint2bits(j0)) ⊦ uint2bits(j / 2 ^ i) = butlastn(i→n(i), uint2bits(j))

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, 2 ^ i ≤ j, 0 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 → uint2bits(j0 / 2 ^ i) = butlastn(i→n(i), uint2bits(j0)) ⊦ uint2bits(j / 2 ^ i) = butlastn(i→n(i), uint2bits(j / 2) + (j % 2 = 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;% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ 2 ^ i ≤ 1 ↔ i < 1;% ⊦ 0 ≤ i → (i < 1 ↔ i = 0);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ 0 ≤ i → ¬ 2 ^ i ≤ 0;% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ butlastn(0, bits) = bits;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: j