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Proof step 10 from proof ls-fin-add (previous step)

i→n(i) = m + 1 ∧ i→n(i) ≠ 0, n→i(m) + 1 < i, 0 ≤ i, 0 < n→i(n) + i, n→i(m) < i, 0 < i ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i→n(i) = m + 1, 0 < n→i(n) + i, n→i(m) < i, n→i(m) + 1 < i, 0 < i ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs