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Proof step 16 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
nσ[n0] = n + 1, # nσ = ∞, stutf(nσ), ∞.n ≤ max∞(λ n0. nσ[n0] = n).n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n0 = 0

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1