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Proof step 9 from proof nat2int-04 (previous step)

n→i(m0) < n→i(n -1) ↔ ¬ n -1 ≤ m0 ⊦ n = 0, n→i(m0) + 1 < n→i(n) ↔ ¬ n ≤ m0 + 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
n = m +1, n→i(m0) < n→i(m) ↔ ¬ m ≤ m0 ⊦ n = 0, n→i(m0) + 1 < n→i(n) ↔ ¬ n ≤ m0 + 1

Rule argument

spec: nat-basic1 inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n -1 ↔ n = m +1 subst: [n, m] → [n, m] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n -1 ↔ n = m +1; (elim) Used terms: n, m