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Proof step 11 from proof make (previous step)

⊦ rational(rational(i, i0).num, rational(i, i0).denom).num = rational(i, i0).num ∧ rational(rational(i, i0).num, rational(i, i0).denom).denom = rational(i, i0).denom, i ≠ 0 ∧ ¬ i0 ≤ 0, ¬ i ≤ 0 ∧ ¬ 0 ≤ i0, ¬ 0 ≤ i ∧ ¬ 0 ≤ i0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

Used simplifier rules: numerator-zero: ⊦ rational(0, j).num = 0;%denominator-zero: ⊦ rational(0, j).denom = 1;