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Proof step 5 from proof less-03 (previous step)

n + n0 < m, n1 < m0 ⊦ m = 0, m0 = n1, m0 = 0, n + n0 + n1 ≠ m + m0 ∧ ¬ m + m0 < n + n0 + n1

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
⊦ m < m0 ∧ n < n0 → (m + n < m0 + n0 ↔ true)

Rule argument

spec: <toplevel> name: l-03 seq: ⊦ m < m0 ∧ n < n0 → (m + n < m0 + n0 ↔ true) subst: [m, m0, n, n0] → [n1, m0, n + n0, m]

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%ls-02: ⊦ m < n → ¬ n < m;%f: ⊦ m < n → n ≠ 0;%l: ⊦ m + n < m + n0 ↔ n < n0;%ref: ⊦ ¬ n < n; Used terms: n1, m0, n + n0, m