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Proof step 65 from proof inv-ls-succ (previous step)

m < m0, n < nσ[m], ⌜m⌝ < # nσ, 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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
m < m0, n < nσ[m], ⌜m⌝ < # nσ, 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 < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic