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Proof step 51 from proof inv-ls-succ (previous step)

max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, (n1 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n1 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n1 - 1)))) ∧ (∀ m. (m ≠ 0 → (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m - 1)))) → ¬ n1 < m), 1 + n < n1, ∞.n ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], stutf(nσ) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 52:
max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, nσ[m] = n1 - 1, 1 + n < n1, ∞.n ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], stutf(nσ), ⌜m⌝ < # nσ, # nσ ≠ ∞ ∨ ¬(m < m0 → nσ[m0] = n1 - 1), ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ n1 < m2 ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1