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Proof step 2 from proof num-neg-03 (previous step)

⊦ ri.num + (~ ri).num = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
ri ≠ 0r ⊦ ∃ i, j. ri = rational(i, j) ∧ ri.num = i ∧ ri.denom = j ∧ 0 < j ∧ gcd(i, j) = 1
Proof step 4:
⊦ ri ≠ 0r, ri.num + (~ ri).num = 0
Proof step 15:
∃ i, j. ri = rational(i, j) ∧ ri.num = i ∧ ri.denom = j ∧ 0 < j ∧ gcd(i, j) = 1 ⊦ ri.num + (~ ri).num = 0

Rule argument

name: exrat-gcd1 subst: [ri] → [ri]

Additional information

Used terms: ri