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Proof step 9 from proof ls-fin-add (previous step)

n→i(i→n(i) - 1) + 1 < i, 0 ≤ i, 0 < n→i(n) + i, n→i(i→n(i) - 1) < i, 0 < i ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
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 ⊦

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [i→n(i), m] flags:

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: i→n(i), m