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Proof step 3 from proof psh-04 (previous step)

m < n -1 ⊦ m + 1 = n, n = 0, n = 1, m ≠ n ∧ ¬ n < m

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ n ≠ 0 → (n0 = n -1 ↔ n = n0 +1)

Rule argument

spec: <toplevel> name: elim seq: ⊦ n ≠ 0 → (n0 = n -1 ↔ n = n0 +1) subst: [n, n0] → [n, n0] flags: <rotated> <all paths>

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%oz: ⊦ m + 1 ≠ 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%one-06: ⊦ 1 = m + 1 ↔ m = 0;%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%f: ⊦ m < n → n ≠ 0;%eq-02: ⊦ m = m + n ↔ n = 0;%l-02: ⊦ ¬ m + n < m;%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1;%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1;%l-05: ⊦ m + m0 < n → ¬ n + n0 < m; Used terms: n, n0