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Proof step 58 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[n2] = nσ[n1] + 1, n1 + m + 1 < n2, ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), nσ[n2 - 1] ≤ nσ[n2], ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. nσ[m] = nσ[n2] → ¬ n2 < m, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n2]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 59:
# nσ = ∞, nσ[n2] = nσ[n1] + 1, n2 = m1 + 1, ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < n2, stutf(nσ), nσ[m1] ≤ nσ[n2], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. nσ[m] = nσ[n2] → ¬ n2 < m, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ nσ[n1 + m + 1] = nσ[n2], n2 = 0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n2, m1]

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ[0] = 0;% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n2, m1