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Proof step 3 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) = rational(i + j, 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ri = rj ↔ ri.num = rj.num ∧ ri.denom = rj.denom
Proof step 5:
⊦ 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

Rule argument

name: Equality rotate: F subst: [rj, ri] → [rational(i + j, 1), 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)]

Additional information

Used terms: rational(i + j, 1), 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)