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Proof step 7 from proof int2bits-one-bit (previous step)

i = 0 ∧ ¬ boolvar ∨ i = ~1 ∧ boolvar ⊦ int2bits(i) = boolvar '

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';%int2bits-mone-02: ⊦ int2bits(~1) = true ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;