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Proof step 3 from proof pe-01 (previous step)

m + n = m1 +1 ∧ m + n ≠ 0, m0 < m ⊦ n0 + m1 = ((n0 + m) + n)-1, m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
(n0 + m) + n = m2 +1 ∧ (n0 + m) + n ≠ 0, m + n = m1 +1 ∧ m + n ≠ 0, m0 < m ⊦ n0 + m1 = m2, m = 0

Rule argument

spec: nat-basic1 inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n -1 ↔ n = m +1 subst: [n, m] → [(n0 + m) + n, m2] flags:

Additional information

Used simplifier rules: z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%n ≠ 0 ⊦ m = n -1 ↔ n = m +1; (elim) Used terms: (n0 + m) + n, m2