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Proof step 8 from proof q-add-01 (previous step)

m0 + m = n ∨ n ≠ m0 ∧ ¬ m0 < n ⊦ (m0 + 1) + m -1 = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

User applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
n ≠ 0 ⊦ m = n -1 ↔ n = m +1
Proof step 10:
m = 0, m0 + m = n ∨ n ≠ m0 ∧ ¬ m0 < n ⊦ (m0 + 1) + m -1 = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

Rule argument

spec: <toplevel> name: elim-pred-c seq: n ≠ 0 ⊦ m = n -1 ↔ n = m +1 subst: [n, m] → [m, m1]

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;%neq-01: ⊦ m ≠ n → (m < n ↔ ¬ n < m);%f: ⊦ m < n → n ≠ 0;%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%one-06: ⊦ 1 = m + 1 ↔ m = 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: m, m1