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Proof step 18 from proof inv-get-inverse-succ (previous step)

∃ n1. 0 < n1 ∧ nσ[n1] = n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n, ∀ n. ∃ m. nσ[m] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# nσ = ∞, stutf(nσ), ∀ n1. ∃ n2. n1 < n2 ∧ nσ[n2] = nσ[n0], ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n0 = 0

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1