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Proof step 1 from proof two-ls (conclusion)

⊦ 0 < j → (j < 2 ↔ j = 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < j, j < 2 ⊦ j = 1

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j;%trans: ⊦ i < j ∧ j < j0 → i < j0;