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Proof step 11 from proof uint2bits-expos0 (previous step)

0 < i → uint2bits(i).rest.first, uint2bits(i) = false ' + bits1, false ∈ uint2bits(i), 0 < i ⊦ uint2bits(i) = false ' + true ' + bits1.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
uint2bits(i) = false ' + bits1, false ∈ uint2bits(i), 0 < i, uint2bits(i).rest.first ⊦ uint2bits(i) = false ' + true ' + bits1.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

Rule argument

Additional information

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