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Proof step 3 from proof less (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ m < m0 ∧ n < n0 → (m + n < m0 + n0 ↔ true)
Proof step 5:
n < m, n < m ∧ n0 < m0 → (n + n0 < m + m0 ↔ true) ⊦ m = 0, n + n0 ≠ m + m0 ∧ ¬ m + m0 < n + n0, m0 ≠ n0 ∧ ¬ n0 < m0

Rule argument

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

Additional information

Used terms: n, m, n0, m0