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Proof step 18 from proof inv-ls-succ (previous step)

nσ[n0] = n + 1, # nσ = ∞, stutf(nσ), ∞.n ≤ max∞(λ n0. nσ[n0] = n).n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1, ∃ m0. n0 < m0 ∧ nσ[n0] < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# nσ = ∞, nσ[n0] = n + 1, n + 1 < nσ[m], n0 < m, ∞.n ≤ max∞(λ n0. nσ[n0] = n).n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1 ⊦ n0 = 0

Rule argument

Additional information

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