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Proof step 3 from proof uint2bits-expos0 (previous step)

uint2bits(i) = boolvar + bits0 ∧ uint2bits(i) ≠ [], 0 < 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 4:
uint2bits(i) = boolvar ' + bits0, 0 < i, boolvar ∈ uint2bits(i) ⊦ uint2bits(i) = false ' + true ' + bits0.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

Rule argument

Additional information

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