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Proof step 22 from proof leninv-succ (previous step)

# nσ = ∞, nσ[m0] = m3 + 1, m < m0, stutf(nσ), nσ[m] ≤ m3, ∀ m0. (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1)) → ¬ nσ[m] + 1 < m0 ⊦ m3 + 1 = nσ[m], ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1), ∃ m1. nσ[m1] = m3 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = nσ[m1])

User applied rule insert lemma

And obtained the following new goals

Proof step 23:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 24:
# nσ = ∞, nσ[m0] = m3 + 1, m < m0, stutf(nσ), nσ[m] ≤ m3, stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ m3 ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = m3), ∀ m0. (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1)) → ¬ nσ[m] + 1 < m0 ⊦ m3 + 1 = nσ[m], ∃ m1. nσ[m1] = m3 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = nσ[m1]), ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 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, m0, m3]

Additional information

Used terms: nσ, m0, m3