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Proof step 5 from proof monls-exists-maxls (previous step)

mon<(nσ), nσ[0] ≤ n, n < nσ[m0 + 1], ⌜m0⌝ + ⌜1⌝ < # nσ, nσ[0] < nσ[m0 + 1], n < nσ[m0], nσ[0] < nσ[m0], ∀ nσ, n. mon<(nσ) ∧ nσ[0] ≤ n ∧ n < nσ[m0] ∧ ⌜m0⌝ < # nσ → (∃ m1. nσ[m1] ≤ n ∧ n < nσ[m1 + 1] ∧ m1 ≤ m0) ⊦ nσ[m0 + 1] = 0, nσ[m0] = 0, ∃ m. nσ[m] ≤ n ∧ n < nσ[m + 1] ∧ m ≤ m0 + 1

User applied rule all left

And obtained the following new goals

Proof step 6:
mon<(nσ), nσ[0] ≤ n, n < nσ[m0 + 1], ⌜m0⌝ + ⌜1⌝ < # nσ, nσ[0] < nσ[m0 + 1], n < nσ[m0], nσ[0] < nσ[m0], mon<(nσ) ∧ nσ[0] ≤ n ∧ n < nσ[m0] ∧ ⌜m0⌝ < # nσ → (∃ m1. nσ[m1] ≤ n ∧ n < nσ[m1 + 1] ∧ m1 ≤ m0) ⊦ nσ[m0 + 1] = 0, nσ[m0] = 0, ∃ m. nσ[m] ≤ n ∧ n < nσ[m + 1] ∧ m ≤ m0 + 1

Rule argument

pos: left 8 terms: nσ, n abort: F discard: T computed: T

Additional information

Used terms: nσ, n