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Proof step 25 from proof leninv-inf (previous step)

∀ n. ∃ m. nσ[m] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], n ≠ 0 → nσ[m] = n - 1 ∧ ¬(m < m0 → nσ[m0] = n - 1) ⊦ n0 < n0 +1 ∧ (∃ m. nσ[m] = n0 +1 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 +1 - 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# nσ = ∞, stutf(nσ), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], n ≠ 0 → nσ[m] = n - 1 ∧ ¬(m < m0 → nσ[m0] = n - 1) ⊦ ∃ m1. nσ[m1] = n0 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = nσ[m1])

Rule argument

Additional information

Used simplifier rules: add-base: ⊦ n + 0 = n; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2