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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
nσ[n] = nσ[n0], # nσ = ∞, n0 < n, m < n, stutf(nσ), n0 < m, nσ[n0] < nσ[m], n0 + 1 < n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[∞.n] = nσ[n0], n0 = 0

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1