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Proof step 21 from proof max-zero-01 (previous step)

1 ≤ m, 3 ≤ m, 6 ≤ m ⊦ maxenc(3, m) = 1 ↔ m = 1 ∨ m = 2, m = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 22:
⊦ m ≤ maxenc(m, n)
Proof step 23:
6 ≤ m, 3 ≤ m, 1 ≤ m, 3 ≤ maxenc(3, m) ⊦ maxenc(3, m) = 1 ↔ m = 1 ∨ m = 2, m = 0

Rule argument

spec: <toplevel> name: max-bigger seq: ⊦ m ≤ maxenc(m, n) subst: [m, n] → [3, m]

Additional information

Used terms: 3, m