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Proof step 75 from proof inv-ls-succ (previous step)

nσ[m] = n + 1, nσ[n0] = n + 1, ⌜m0⌝ < # nσ, stutf(nσ), ⌜m⌝ < # nσ, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m < m0, m < n0, ⌜n0⌝ < # nσ, n + 1 < nσ[m0], n < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ 2 + n < m2 ⊦ nσ[m0] = n + 1

User applied rule all left

And obtained the following new goals

Proof step 76:
nσ[m] = n + 1, nσ[n0] = n + 1, ⌜m0⌝ < # nσ, stutf(nσ), ⌜m⌝ < # nσ, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m < m0, m < n0, ⌜n0⌝ < # nσ, n + 1 < nσ[m0], n < nσ[m0], ∃ n1. m0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ 2 + n < m2 ⊦ nσ[m0] = n + 1

Rule argument

pos: left 12 terms: m0 abort: F discard: T computed: F

Additional information

Used terms: m0