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Proof step 2 from proof fst-nle (previous step)

m0 ≠ 0, m0 ≠ 1 ⊦ n + m * m0 ≤ m.1 ↔ m = 0 ∧ n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ m.1 ≤ m
Proof step 4:
m.1 ≤ m, m0 ≠ 0, m0 ≠ 1 ⊦ n + m * m0 ≤ m.1 ↔ m = 0 ∧ n = 0

Rule argument

spec: nat-pair inst: name: fst-ls seq: ⊦ m.1 ≤ m subst: [m] → [m] flags: current, []

Additional information

Used terms: m