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Proof step 112 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, ∀ n1. ¬ ⌜n1⌝ < # nσ ∨ nσ[n1] ≠ n ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 113:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 114:
stutf(nσ) ∧ ⌜n0⌝ < # nσ ∧ n ≤ nσ[n0] → (∃ m0. m0 ≤ n0 ∧ nσ[m0] = n), 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, ∀ n1. ¬ ⌜n1⌝ < # nσ ∨ nσ[n1] ≠ n ⊦

Rule argument

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

Additional information

Used terms: nσ, n0, n