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Proof step 116 from proof inv-ls-succ (previous step)

nσ[n0] = n + 1, leninv(nσ) = ∞, ⌜n0⌝ < # nσ, stutf(nσ), n0 ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = n + 1 → ¬ n0 < m ⊦ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 117:
max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ⌜n1⌝ , nσ[n0] = n + 1, leninv(nσ) = ∞, ⌜n0⌝ < # nσ, stutf(nσ), n0 ≤ n1, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = n + 1 → ¬ n0 < m ⊦ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n) = ∞

Rule argument

spec: nat-inf-basic inst: name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n), m0] flags:

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n), m0