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Proof step 1 from proof exrat-01 (conclusion)

⊦ ∃ i, j. ri = rational(i, j) ∧ 0 < j

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ ∃ i, j. ri = rational(i, j) ∧ ri.num = i ∧ ri.denom = j ∧ 0 < j
Proof step 3:
∃ i, j. ri = rational(i, j) ∧ ri.num = i ∧ ri.denom = j ∧ 0 < j ⊦ ∃ i, j. ri = rational(i, j) ∧ 0 < j

Rule argument

name: exrat subst: [ri] → [ri]

Additional information

Used terms: ri