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Proof step 140 from proof inv-ls-succ (previous step)

nσ[n0] = n + 1, leninv(nσ) = ⌜n1⌝ , ⌜n0⌝ < # nσ, 1 + n < n1, n + 1 < n1, stutf(nσ), n0 ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = n + 1 → ¬ n0 < m, ∀ n2. ¬ ⌜n2⌝ < # nσ ∨ nσ[n2] ≠ n ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 141:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 142:
nσ[n0] = n + 1, leninv(nσ) = ⌜n1⌝ , n + 1 < n1, 1 + n < n1, ⌜n0⌝ < # nσ, stutf(nσ), n0 ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], stutf(nσ) ∧ ⌜n0⌝ < # nσ ∧ n ≤ nσ[n0] → (∃ m0. m0 ≤ n0 ∧ nσ[m0] = n), ∀ n2. ¬ ⌜n2⌝ < # nσ ∨ nσ[n2] ≠ n, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = n + 1 → ¬ n0 < m ⊦

Rule argument

spec: <toplevel> name: stutf-alltaken seq: ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n) subst: [nσ, m, n] → [nσ, n0, n]

Additional information

Used terms: nσ, n0, n