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Proof step 73 from proof inv-ls-succ (previous step)

nσ[m] = n + 1, nσ[n0] = n + 1, m < n0, ⌜n0⌝ < # nσ, m < m0, ⌜m⌝ < # nσ, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, mon≤(nσ) ∧ m ≤ m0 ∧ ⌜m0⌝ < # nσ → nσ[m] ≤ nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ m2. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m2 - 1))) → ¬ 2 + n < m2 ⊦ nσ[m0] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 74:
nσ[m] = n + 1, nσ[n0] = n + 1, m < n0, ⌜n0⌝ < # nσ, m < m0, ⌜m⌝ < # nσ, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ⌜m0⌝ < # nσ → n < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ m2. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m2 - 1))) → ¬ 2 + n < m2 ⊦ nσ[m0] = n + 1

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n → m ≤ n; (s)