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Proof step 2 from proof neq-02 (previous step)

m + n0 + 1 = n + m0, m < m0, n0 < n ⊦ m0 = 0, n = 0

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: nat-basic2 inst: name: less-05 seq: m < m0, ¬ n0 < n ⊦ m + n < m0 + n0 subst: [m, m0, n0, n] → [n0, n, m0, m + 1] flags: current, []

Additional information

Used simplifier rules: s: ⊦ m < n + 1 ↔ ¬ n < m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: n0, n, m0, m + 1