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Proof step 19 from proof inv-get-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# nσ = ∞, nσ[n0] < nσ[m], n0 < m, stutf(nσ), ∀ n1. ∃ n2. n1 < n2 ∧ nσ[n2] = nσ[n0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[∞.n] = nσ[n0], n0 = 0

Rule argument

Additional information

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