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Proof step 4 from proof elim-sub (previous step)

0 < j ⊦ ri0 = rational(i, j) - rj ↔ rational(i, j) = ri0 + rj

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ ∃ i, j. ri = rational(i, j) ∧ 0 < j
Proof step 6:
∃ i, j. ri0 = rational(i, j) ∧ 0 < j, 0 < j ⊦ ri0 = rational(i, j) - rj ↔ rational(i, j) = ri0 + rj

Rule argument

name: exrat-01 subst: [ri] → [ri0]

Additional information

Used terms: ri0