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Proof step 4 from proof uint2bits-falsein (previous step)

bitsinvert(uint2bits(i)) ≠ [], 1 < i, 0 < i, 0 ≤ i ⊦ bitsinvert(uint2bits(i)).last ∨ false ∈ bitsinvert(uint2bits(i)).butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
1 < i, 0 < i, 0 ≤ i ⊦ bitsinvert(uint2bits(i)).last, false ∈ bitsinvert(uint2bits(i)).butlast

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ bitsinvert(bits) = [] ↔ bits = [];