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Proof step 11 from proof abs-ls-same-01 (previous step)

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

User applied rule insert equation

And obtained the following new goals

Proof step 12:
j = abs(j) ⊦ abs(i) * abs(j) < i ↔ ¬ i ≤ 0 ∧ ¬ 0 < abs(j) ∨ ¬ 0 ≤ i ∧ ¬ ~1 ≤ abs(j)

Rule argument

pos: left 1 flags:

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