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Proof step 1 from proof int2bits-falsein-03 (conclusion)

⊦ ¬ false ∈ bits ∧ ¬(∃ i. j = 2 ^ i - 1) → int2bits(j) ≠ boolvar ' + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ i - j = i0 ↔ i = i0 + j;