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

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) ⊦ ∃ m1. nσ[m1] ≤ n ∧ n < nσ[m1 + 1] ∧ m1 ≤ m0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ m +1 = m + 1; (s)