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Proof step 5 from proof snd-nle (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m.2 ≤ m, n + m * m0 ≤ m.2, n + m * m0 ≤ m ⊦ m0 = 0, m = 0 ∧ n = 0

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat