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Proof step 3 from proof monls-exists-maxleq (previous step)

n = nσ[m], nσ[0] < nσ[m], ⌜m⌝ < # nσ, n ≤ nσ[m], nσ[0] < n, mon<(nσ) ⊦ n = 0, nσ[m] = 0, ∃ m0. nσ[m0] < n ∧ n ≤ nσ[m0 + 1] ∧ m0 ≤ m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⌜m⌝ < # nσ, nσ[0] < nσ[m], mon<(nσ) ⊦ nσ[m] = 0, ∃ m0. nσ[m0] < nσ[m] ∧ nσ[m] ≤ nσ[m0 + 1] ∧ m0 ≤ m

Rule argument

Additional information

Used simplifier rules: ler: ⊦ m ≤ m; from specification nat