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Proof step 14 from proof uint2bits-inj (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
uint2bits(i / 2) = uint2bits(j / 2), i % 2 = 1 ↔ j % 2 = 1, 0 < j, 0 < i, 0 ≤ j, 0 ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → uint2bits(i0) = uint2bits(j) ∧ 0 ≤ i0 ∧ 0 ≤ j → i0 = j ⊦ i = j, i = 1, j = 1

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ bits + boolvar ' = bits0 + boolvar0 ' ↔ bits = bits0 ∧ (boolvar ↔ boolvar0);% ⊦ bits + boolvar = bits + boolvar ';