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Proof step 53 from proof eq-leninv (previous step)

m + 1 = nσ[n], # nσ = ∞, stutf(nσ), ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ ∃ m0. nσ[m0] = m ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

User applied rule insert lemma

And obtained the following new goals

Proof step 54:
⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n < nσ[m] → (∃ m0. m0 < m ∧ nσ[m0] = n ∧ nσ[m0 + 1] = n + 1)
Proof step 55:
stutf(nσ) ∧ ⌜n⌝ < # nσ ∧ m < nσ[n] → (∃ m0. m0 < n ∧ nσ[m0] = m ∧ nσ[m0 + 1] = m + 1), m + 1 = nσ[n], # nσ = ∞, stutf(nσ), ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ ∃ m0. nσ[m0] = m ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

Rule argument

spec: inst: name: stutf-alltakenmax seq: ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n < nσ[m] → (∃ m0. m0 < m ∧ nσ[m0] = n ∧ nσ[m0 + 1] = n + 1) subst: [nσ, m, n] → [nσ, n, m] flags: current, []

Additional information

Used terms: nσ, n, m