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Proof step 15 from proof ls-even-01 (previous step)

i % 2 = 0, j % 2 = 0 ⊦ i / 2 < j / 2 ↔ ¬ j ≤ i, ¬ i < 0 ∧ ¬ j < 0

User applied rule case distinction

And obtained the following new goals

Proof step 16:
j % 2 = 0, i % 2 = 0, i < 0 ⊦ i / 2 < j / 2 ↔ ¬ j ≤ i
Proof step 31:
j % 2 = 0, i % 2 = 0, j < 0 ⊦ i / 2 < j / 2 ↔ ¬ j ≤ i

Rule argument

pos: right 2

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