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Proof step 3 from proof exrat-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
rational(i, j).num = i, rational(i, j).denom = j, 0 < j ⊦ ∃ i0, j0. rational(i, j) = rational(i0, j0) ∧ ¬ j0 ≤ 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;