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Proof step 1 from proof int2bits-first-05 (conclusion)

⊦ int2bits(i) = boolvar ' + bits ∧ 0 ≤ i → ¬ boolvar

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: int2bits-first-04: ⊦ 0 ≤ j → int2bits(j) ≠ true ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;