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Proof step 2 from proof uint2bits-not-one (previous step)

uint2bits(i) = true ', 0 ≤ i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < i ∧ i ≠ 1 → uint2bits(i) = uint2bits(i / 2) + (i % 2 = 1)

Rule argument

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

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;%uint2bits-zero: ⊦ uint2bits(0) = false ';%uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ boolvar ' = bits + boolvar0 ' ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ bits + boolvar = bits + boolvar ';% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: i