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

0 < i → uint2bits(i).rest.first, uint2bits(i) = false ' + false ' + bits, 0 ≤ i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
uint2bits(i) = false ' + false ' + bits, uint2bits(i).rest.first, 0 < i ⊦ 0 = i

Rule argument

Additional information

Used simplifier rules: uint2bits-zero: ⊦ uint2bits(0) = false ';% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;