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Proof step 34 from proof inv-get-inverse (previous step)

# nσ = ∞, nσ[n] < nσ[m], n < m, stutf(nσ), ∀ n1. nσ[n1] ≠ nσ[n] + 1 ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 35:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 36:
stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ nσ[n] + 1 ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = nσ[n] + 1), # nσ = ∞, nσ[n] < nσ[m], n < m, stutf(nσ), ∀ n1. nσ[n1] ≠ nσ[n] + 1 ⊦

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, m, nσ[n] + 1] flags: current, []

Additional information

Used terms: nσ, m, nσ[n] + 1