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Proof step 4 from proof abs-pos (previous step)

0 < j ⊦ rational(abs(rational(i, j).num), rational(i, j).denom) = rational(abs(i), j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ri = rj ↔ ri.num = rj.num ∧ ri.denom = rj.denom
Proof step 6:
0 < j ⊦ rational(abs(rational(i, j).num), rational(i, j).denom).num = rational(abs(i), j).num ∧ rational(abs(rational(i, j).num), rational(i, j).denom).denom = rational(abs(i), j).denom

Rule argument

name: Equality rotate: F subst: [rj, ri] → [rational(abs(i), j), rational(abs(rational(i, j).num), rational(i, j).denom)]

Additional information

Used terms: rational(abs(i), j), rational(abs(rational(i, j).num), rational(i, j).denom)