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Proof step 4 from proof monls-exists-maxls (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 5:
⌜m0⌝ + ⌜1⌝ < # nσ, n < nσ[m0 + 1], nσ[0] < nσ[m0 + 1], mon<(nσ), nσ[0] ≤ n, ∀ nσ, n. ⌜m0⌝ < # nσ ∧ n < nσ[m0] ∧ mon<(nσ) ∧ nσ[0] ≤ n → (∃ m1. nσ[m1] ≤ n ∧ n < nσ[m1 + 1] ∧ m1 ≤ m0) ⊦ ∃ m. nσ[m] ≤ n ∧ n < nσ[m + 1] ∧ m ≤ m0 + 1, nσ[m0] ≤ n ∧ n < nσ[m0 + 1] ∧ m0 ≤ m0 + 1

Rule argument

pos: right 1 terms: m0 abort: F discard: F computed: T

Additional information

Used terms: m0