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Proof step 66 from proof inv-ls-succ (previous step)

stutf(nσ), ⌜m⌝ < # nσ, n < nσ[m], ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m < m0, ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ nσ[m] + 1 < m2, ∃ n1. m < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = nσ[m]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 67:
nσ[n0] = n + 1, stutf(nσ), ⌜m⌝ < # nσ, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m < m0, m < n0, ⌜n0⌝ < # nσ, n < nσ[m], ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ nσ[m] + 1 < m2, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = nσ[m]

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat