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Proof step 7 from proof add-asso (previous step)

0 < j, 0 < j0 ⊦ (rational(i, j) + rational(i0, j0)) + rj0 = rational(i, j) + rational(i0, j0) + rj0

User applied rule insert lemma

And obtained the following new goals

Proof step 8:
⊦ ∃ i, j. ri = rational(i, j) ∧ 0 < j
Proof step 9:
∃ i, j. rj0 = rational(i, j) ∧ 0 < j, 0 < j, 0 < j0 ⊦ (rational(i, j) + rational(i0, j0)) + rj0 = rational(i, j) + rational(i0, j0) + rj0

Rule argument

name: exrat-01 subst: [ri] → [rj0]

Additional information

Used terms: rj0