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Proof step 3 from proof le-same-04 (previous step)

abs(j) = 1 * abs(j) ⊦ i * abs(j) ≤ abs(j) ↔ j = 0 ∨ i ≤ 1

User applied rule insert equation

And obtained the following new goals

Proof step 4:
⊦ i * abs(j) ≤ 1 * abs(j) ↔ j = 0 ∨ i ≤ 1

Rule argument

pos: left 1 flags: discard, [[2,1,1,2]]

Additional information