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Proof step 20 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# nσ = ∞, stutf(nσ), n0 < m, nσ[n0] < nσ[m], ∀ n1. ∃ n2. n1 < n2 ∧ nσ[n2] = nσ[n0], ∀ n. ∃ m. nσ[m] = n ⊦ n0 = 0, m = 0, nσ[m] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1