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Proof step 21 from proof int2bits-truein-01 (previous step)

∃ bits. uint2bits(j) = false ' + true ' + bits, 0 < j ⊦ j = 0, true ∈ uint2bits(j).rest, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
uint2bits(j) = false ' + true ' + bits, false ∈ uint2bits(j), 0 < j ⊦ j = 0, true ∈ uint2bits(j).rest, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

Rule argument

Additional information

Used simplifier rules: uint2bits-onein: ⊦ 0 < j → true ∈ uint2bits(j);% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)