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Proof step 18 from proof stutf-maxinf-is-fin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
nσ[n1] = nσ[m0], # nσ = ∞, m0 < n1, n0 < n1, nσ[m0] < nσ[n0], m0 ≤ n, nσ[m0] ≤ nσ[n], stutf(nσ), m0 ≤ n0 ⊦ n1 = 0, nσ[n0] = 0, nσ[n0] = nσ[m0]

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1