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Proof step 4 from proof uint2bits-inj (previous step)

uint2bits(i) = uint2bits(j), 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → uint2bits(i0) = uint2bits(j) ∧ 0 ≤ i0 ∧ 0 ≤ j → i0 = j ⊦ i = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i ∧ i ≠ 1 → uint2bits(i) = uint2bits(i / 2) + (i % 2 = 1)
Proof step 6:
¬(0 < i ∧ i ≠ 1), uint2bits(i) = uint2bits(j), 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → uint2bits(i0) = uint2bits(j) ∧ 0 ≤ i0 ∧ 0 ≤ j → i0 = j ⊦ i = j
Proof step 9:
0 < i ∧ i ≠ 1, uint2bits(i / 2) + (i % 2 = 1) = uint2bits(j), 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → uint2bits(i0) = uint2bits(j) ∧ 0 ≤ i0 ∧ 0 ≤ j → i0 = j ⊦ i = j

Rule argument

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

Additional information

Used terms: i