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Proof step 92 from proof inv-ls-succ (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 93:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 94:
⌜m⌝ < # nσ, n < nσ[m], m < m0, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n + 1 ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n + 1), ∀ m0. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m0 - 1))) → ¬ nσ[m] + 1 < m0, ∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ[n0] ≠ n + 1 ⊦ nσ[m0] = nσ[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, m, n + 1]

Additional information

Used terms: nσ, m, n + 1