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Proof step 35 from proof inv-ls-succ (previous step)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. nσ[n] ≠ (n2 + 1) + 1, ∀ n0, nσ. n0 < n2 +1 → (∀ n1. nσ[n1] ≠ n0 + 1) ∧ # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦

User applied rule apply induction

And obtained the following new goals

Proof step 36:
# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. nσ[n] ≠ (n2 + 1) + 1, ∀ n0, nσ. n0 < n2 +1 → (∀ n1. nσ[n1] ≠ n0 + 1) ∧ # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦ (∀ n1. nσ[n1] ≠ n2 + 1) ∧ # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

Rule argument

subst: [n0, nσ] → [n2, nσ]

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: n2, nσ