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Proof step 4 from proof int2bits-falsein-03 (previous step)

false ∈ int2bits(j).rest ↔ ¬(∃ i. j = 2 ^ i - 1), int2bits(j) = boolvar ' + bits, boolvar ∈ int2bits(j) ⊦ false ∈ bits, ∃ i. 2 ^ i = j + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
int2bits(j) = boolvar ' + bits, false ∈ int2bits(j).rest ↔ (∀ i. 2 ^ i ≠ j + 1), boolvar ∈ int2bits(j) ⊦ false ∈ bits, ∃ i. 2 ^ i = j + 1

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;