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Proof step 16 from proof eq-one (previous step)

⊦ ¬ i ≤ 0 ∧ ¬ 0 ≤ j, ¬ 0 ≤ i ∧ ¬ j ≤ 0, i * j = 1 ↔ abs(j) = 1 ∧ i = j, ¬ i ≤ 0 ∧ 1 < j

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 17:
i ≤ 0 ⊦ ¬ i ≤ 0 ∧ ¬ 0 ≤ j, ¬ 0 ≤ i ∧ ¬ j ≤ 0, i * j = 1 ↔ abs(j) = 1 ∧ i = j, ¬ i ≤ 0 ∧ 1 < j
Proof step 23:
⊦ i ≤ 0, ¬ i ≤ 0 ∧ ¬ 0 ≤ j, ¬ 0 ≤ i ∧ ¬ j ≤ 0, i * j = 1 ↔ abs(j) = 1 ∧ i = j, ¬ i ≤ 0 ∧ 1 < j

Rule argument

fma: i ≤ 0

Additional information

Used formulas: i ≤ 0