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Proof step 12 from proof uint2bits-inj (previous step)

¬(0 < j ∧ j ≠ 1), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 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);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;