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Proof step 1 from proof add-asso (conclusion)

⊦ (ri + rj) + rj0 = ri + rj + rj0

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ ∃ i, j. ri = rational(i, j) ∧ 0 < j
Proof step 3:
∃ i, j. ri = rational(i, j) ∧ 0 < j ⊦ (ri + rj) + rj0 = ri + rj + rj0

Rule argument

name: exrat-01 subst: [ri] → [ri]

Additional information

Used terms: ri