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Proof step 4 from proof one-02 (previous step)

j = i +1, 0 < j, ∀ j0. i→n(j0) < i→n(j) → true → 0 < j0 → 1 ^ j0 = 1 ⊦ 1 ^ i = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i, ∀ j0. i→n(j0) < i→n(i +1) → true → 0 < j0 → 1 ^ j0 = 1 ⊦ 1 ^ i = 1

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2