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Proof step 5 from proof int-add (previous step)

⊦ rational(rational(i, 1).num * rational(j, 1).denom + rational(i, 1).denom * rational(j, 1).num, rational(i, 1).denom * rational(j, 1).denom).num = rational(i + j, 1).num ∧ rational(rational(i, 1).num * rational(j, 1).denom + rational(i, 1).denom * rational(j, 1).num, rational(i, 1).denom * rational(j, 1).denom) .denom = rational(i + j, 1).denom

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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