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Proof step 75 from proof inv-get-inverse (previous step)

∀ n0. m < n0 ∨ nσ[n0] ≠ n, # nσ = ⌜m⌝ + ⌜1⌝ , n ≤ nσ[m], stutf(nσ) ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 76:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 77:
stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n), ∀ n0. m < n0 ∨ nσ[n0] ≠ n, # nσ = ⌜m⌝ + ⌜1⌝ , n ≤ nσ[m], stutf(nσ) ⊦

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, n] flags: current, []

Additional information

Used terms: nσ, m, n