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Proof step 62 from proof inv-get-inverse (previous step)

# nσ = ∞, stutf(nσ), n ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∀ n0. nσ[n0] ≠ n ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 63:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 64:
# nσ = ∞, stutf(nσ), n ≤ nσ[m], stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n), ∀ n0. nσ[n0] ≠ n, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 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, m, n]

Additional information

Used terms: nσ, m, n