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Proof step 115 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[n1] < nσ[m1], n1 + m + 1 < ∞.n, stutf(nσ), ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, (∀ n. ∃ n0. n < n0 ∧ nσ[n0] = nσ[n1] + 1) ∨ (∀ n. nσ[n] ≠ nσ[n1] + 1) ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 116:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 117:
# nσ = ∞, n1 + m + 1 < ∞.n, nσ[n1] < nσ[m1], stutf(nσ), stutf(nσ) ∧ ⌜m1⌝ < # nσ ∧ nσ[n1] + 1 ≤ nσ[m1] → (∃ m0. m0 ≤ m1 ∧ nσ[m0] = nσ[n1] + 1), ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, (∀ n. ∃ n0. n < n0 ∧ nσ[n0] = nσ[n1] + 1) ∨ (∀ n. nσ[n] ≠ nσ[n1] + 1) ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

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, m1, nσ[n1] + 1]

Additional information

Used terms: nσ, m1, nσ[n1] + 1