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Proof step 13 from proof div-neg-01 (previous step)

rj.num = 0 ⊦ rj = rational(0, 1), rational(ri.num, 0) = ~(ri / rj)

User applied rule insert lemma

And obtained the following new goals

Proof step 14:
⊦ ∃ i, j. ri = rational(i, j) ∧ ri.num = i ∧ ri.denom = j ∧ 0 < j

Rule argument

name: exrat subst: [ri] → [rj]

Additional information

Used simplifier rules: num-zero-denum-one: ⊦ rj.num = 0 → rj.denom = 1;%num-zero: ⊦ rational(i, j).num = 0 ↔ i = 0;%numerator-zero: ⊦ rational(0, j).num = 0;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) Used terms: rj