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Proof step 41 from proof leninv-succ (previous step)

# nσ = ∞, nσ[m2] = m3 + 1, nσ[m0] = nσ[m1] + 1, nσ[m1 + 1] = nσ[m0], nσ[m] = nσ[m1], nσ[m1] < m3, m < m0, m1 < m0, m1 < m2, m < m2, stutf(nσ), nσ[m0] ≤ m3, m ≤ m1 ⊦ m3 + 1 = nσ[m0], ∃ m4. nσ[m4] = m3 ∧ ¬(∀ m. m4 < m → nσ[m] = nσ[m4])

User applied rule insert lemma

And obtained the following new goals

Proof step 42:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n)
Proof step 43:
# nσ = ∞, nσ[m] = nσ[m1], nσ[m1 + 1] = nσ[m0], nσ[m0] = nσ[m1] + 1, nσ[m2] = m3 + 1, m < m2, m1 < m2, m1 < m0, m < m0, nσ[m1] < m3, stutf(nσ), m ≤ m1, nσ[m0] ≤ m3, stutf(nσ) ∧ ⌜m2⌝ < # nσ ∧ m3 ≤ nσ[m2] → (∃ m0. m0 ≤ m2 ∧ nσ[m0] = m3) ⊦ m3 + 1 = nσ[m0], ∃ m4. nσ[m4] = m3 ∧ ¬(∀ m. m4 < m → nσ[m] = nσ[m4])

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, m2, m3]

Additional information

Used terms: nσ, m2, m3