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Proof step 5 from proof two-ls (previous step)

0 < j, j +1 = 2 ∨ j +1 < 2 ⊦ j = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, j < 1 ⊦ j = 1

Rule argument

Additional information

Used simplifier rules: numeral: ⊦ i +1 = (* j) ↔ i = j -1;%numeral-01: ⊦ i +1 < (* j) ↔ i < j -1;