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Proof step 6 from proof uint2bits-expos0 (previous step)

0 ≤ i → ¬ uint2bits(i).first, uint2bits(i) = boolvar ' + bits0, 0 < i, boolvar ∈ uint2bits(i) ⊦ uint2bits(i) = false ' + true ' + bits0.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
uint2bits(i) = boolvar ' + bits0, 0 < i, boolvar ∈ uint2bits(i) ⊦ uint2bits(i) = false ' + true ' + bits0.rest, uint2bits(i).first, ∃ bits. uint2bits(i) = false ' + true ' + bits

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;