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Proof step 67 from proof inv-ls-succ (previous step)

⌜m⌝ < # nσ, n < nσ[m], m < m0, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ∀ m0. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m0 - 1))) → ¬ nσ[m] + 1 < m0, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∃ n1. m < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = nσ[m]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 68:
nσ[n0] = n + 1, m < n0, ⌜n0⌝ < # nσ, m < m0, ⌜m⌝ < # nσ, n < nσ[m], stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ m0. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ nσ[m0] = nσ[m]

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat