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Proof step 93 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 94:
nσ[m1] = n + 1, stutf(nσ), ⌜m⌝ < # nσ, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m < m0, m1 ≤ m, m1 < m0, n < nσ[m], ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ nσ[m] + 1 < m2, ∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ[n0] ≠ n + 1 ⊦ nσ[m0] = nσ[m]

Rule argument

Additional information

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