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Proof step 2 from proof fle (previous step)

m < n, n ≤ n0, n0 ≤ m ⊦ n = 0

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ m < n ∧ ¬ n0 < m → ¬ n0 < n
Proof step 4:
¬ m < n0 ∧ ¬ n < m → ¬ n < n0, m < n, n ≤ n0, n0 ≤ m ⊦ n = 0

Rule argument

spec: nat-basic1 inst: name: lf-07 seq: ⊦ ¬ m < n ∧ ¬ n0 < m → ¬ n0 < n subst: [m, n, n0] → [m, n0, n] flags:

Additional information

Used terms: m, n0, n