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Proof step 2 from proof one-03 (previous step)

0 < i, 0 ≤ j ⊦ 1 = i + j ↔ i = 1 ∧ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
1 = i + j, 0 < i ⊦ j < 0, i = 1 ∧ j = 0

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ i ≤ j ↔ ¬ j < i;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2