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Proof step 53 from proof inv-ls-succ (previous step)

n1 = m3 + 1 ∧ n1 ≠ 0, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, nσ[m] = m3, 1 + n < n1, ∞.n ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], stutf(nσ), ⌜m⌝ < # nσ, # nσ ≠ ∞ ∨ ¬(m < m0 → nσ[m0] = m3), ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ n1 < m2 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 54:
max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, n < nσ[m], ⌜m⌝ < # nσ, stutf(nσ), ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ nσ[m] + 1 < m2, # nσ ≠ ∞ ∨ ¬(m < m0 → nσ[m0] = nσ[m]) ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%get-infto: m < n ⊦ (nnf to n)[m] = nnf(m); from specification stream[natstream]%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2