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Proof step 57 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
n < nσ[m], ⌜m⌝ < # nσ, 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, # nσ ≠ ∞ ∨ ¬(m < m0 → nσ[m0] = nσ[m]), (∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1) ∨ (∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ[n0] ≠ n + 1) ⊦

Proof step 57 from proof inv-ls-succ (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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