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Proof step 2 from proof monls-exists-maxleq (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 3:
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
Proof step 10:
nσ[0] < nσ[m], ⌜m⌝ < # nσ, n ≤ nσ[m], nσ[0] < n, mon<(nσ) ⊦ n = nσ[m], n = 0, nσ[m] = 0, ∃ m0. nσ[m0] < n ∧ n ≤ nσ[m0 + 1] ∧ m0 ≤ m

Rule argument

fma: n = nσ[m]

Additional information

Used formulas: n = nσ[m]