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Proof step 4 from proof add-asso (previous step)

0 < j ⊦ (rational(i, j) + rj) + rj0 = rational(i, j) + rj + rj0

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. rj = rational(i, j) ∧ 0 < j, 0 < j ⊦ (rational(i, j) + rj) + rj0 = rational(i, j) + rj + rj0

Rule argument

name: exrat-01 subst: [ri] → [rj]

Additional information

Used terms: rj