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Proof step 3 from proof int2bits-first-02 (previous step)

int2bits(j) = boolvar + bits ∧ int2bits(j) ≠ [], 0 ≤ j, boolvar ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
int2bits(j) = true ' + bits, 0 ≤ j, true ∈ int2bits(j) ⊦

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ boolvar + bits = boolvar ' + bits;