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Proof step 53 from proof inv-get-inverse-between (previous step)

nσ[n1] = n ∧ (∀ m. nσ[m] = n → ¬ n1 < m), # nσ = ∞, nσ[n2] = n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), n1 + m + 1 < n2, ∀ m. nσ[m] = n + 1 → ¬ n2 < m, ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n2 = 0, nσ[n1 + m + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 54:
# nσ = ∞, nσ[n2] = nσ[n1] + 1, ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), n1 + m + 1 < n2, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. nσ[m] = nσ[n2] → ¬ n2 < m, ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n2 = 0, nσ[n1 + m + 1] = nσ[n2]

Proof step 53 from proof inv-get-inverse-between (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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