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Proof step 13 from proof uint2bits-inj (previous step)

uint2bits(i / 2) + (i % 2 = 1)' = uint2bits(j), 0 < i, 0 ≤ j, 0 ≤ i, (i % 2 = 1) ∈ uint2bits(j), ∀ i0, j. i→n(i0) < i→n(i) → uint2bits(i0) = uint2bits(j) ∧ 0 ≤ i0 ∧ 0 ≤ j → i0 = j ⊦ i = j, i = 1, j ≠ 1 ∧ 0 ≠ j

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;%uint2bits-zero-01: 0 ≤ i ⊦ uint2bits(i) = false ' ↔ i = 0;%uint2bits-zero: ⊦ uint2bits(0) = false ';%uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ boolvar ' = bits + boolvar0 ' ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';%⊦ bits + boolvar ' = bits0 + boolvar0 ' ↔ bits = bits0 ∧ (boolvar ↔ boolvar0);% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ 0 < i → (i / 2 = 0 ↔ i = 1);