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Proof step 13 from proof ls-fin-add (previous step)

i→n(i0) - 1 = m + 1, 1 < i0, 0 < i0, 1 < n→i(n) + i0, 0 < n→i(n) + i0, 2 + n→i(m) < i0 ⊦ i0 = 1, i0 = 0, ∃ m0, n0. i→n(i0) = m0 + 1 ∧ 0 < n→i(n0) + i0 ∧ n→i(m0) < i0 ∧ n→i(m0) + 1 < i0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
i→n(i0) = m1 + 1 ∧ i→n(i0) ≠ 0, m1 = m + 1, 1 < i0, 0 < i0, 1 < n→i(n) + i0, 0 < n→i(n) + i0, 2 + n→i(m) < i0 ⊦ i0 = 1, i0 = 0, ∃ m0, n0. i→n(i0) = m0 + 1 ∧ 0 < n→i(n0) + i0 ∧ n→i(m0) < i0 ∧ n→i(m0) + 1 < i0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [i→n(i0), m1] flags:

Additional information

Used simplifier rules: abs-zero: ⊦ i→n(i) = 0 ↔ i = 0;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: i→n(i0), m1