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Proof step 31 from proof inv-get-inverse-succ (previous step)

(∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n) ∨ (∀ n0. nσ[n0] ≠ n), # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∀ n. ∃ m. nσ[m] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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