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Proof step 3 from proof uint2uint-pos (previous step)

0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → bits2uint(uint2bits(i0)) = i0 ⊦ bits2uint(uint2bits(i)) = i

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 → bits2uint(uint2bits(i0)) = i0 ⊦ bits2uint(uint2bits(i / 2) + (i % 2 = 1)) = i

Rule argument

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

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;%bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);%bits2uint-one-03: ⊦ bits2uint(true ') = 1;%uint2bits-zero: ⊦ uint2bits(0) = false ';%bits2uint-one-02: ⊦ bits2uint(false ') = 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i ≠ j → (i ≤ j ↔ i < j); Used terms: i