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Proof step 3 from proof snd-nle (previous step)

n + m * m0 ≤ m.2 ⊦ m0 = 0, m = 0 ∧ n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ m.2 ≤ m
Proof step 5:
m.2 ≤ m, n + m * m0 ≤ m.2 ⊦ m0 = 0, m = 0 ∧ n = 0

Rule argument

spec: nat-pair inst: name: snd-le seq: ⊦ m.2 ≤ m subst: [m] → [m] flags: current, []

Additional information

Used terms: m