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

# nσ = ∞, stutf(nσ), m < m0, nσ[m] < nσ[m0] ⊦ m0 = 0, nσ[m0] = 0, ∃ m0. nσ[m0] = nσ[m] + 1

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) ∧ ⌜m0⌝ < # nσ ∧ nσ[m] + 1 ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = nσ[m] + 1), # nσ = ∞, stutf(nσ), m < m0, nσ[m] < nσ[m0] ⊦ m0 = 0, nσ[m0] = 0, ∃ m0. nσ[m0] = nσ[m] + 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, m0, nσ[m] + 1] flags: current, []

Additional information

Used terms: nσ, m0, nσ[m] + 1