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Proof step 80 from proof inv-ls-succ (previous step)

nσ[m] = n + 1, nσ[n0] = n + 1, m < m0, n0 < m0, stutf(nσ), ⌜m⌝ < # nσ, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m < n0, ⌜n0⌝ < # nσ, m + 1 < 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, ⌜m0⌝ < # nσ

User applied rule all left

And closed the goal!

Rule argument

pos: left 11 terms: m0 abort: F discard: T computed: F

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: m0