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Proof step 1 from proof int2bits-one-bit (conclusion)

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

User applied rule case distinction

And obtained the following new goals

Proof step 2:
int2bits(i) = boolvar ' ⊦ i = 0 ∧ ¬ boolvar ∨ i = ~1 ∧ boolvar
Proof step 7:
i = 0 ∧ ¬ boolvar ∨ i = ~1 ∧ boolvar ⊦ int2bits(i) = boolvar '

Rule argument

pos: right 1

Additional information