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Proof step 136 from proof inv-ls-succ (previous step)

∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n, nσ[n0] = n + 1, leninv(nσ) = ⌜n1⌝ , n + 1 < n1, n0 ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], 1 + n < n1, stutf(nσ), ⌜n0⌝ < # nσ, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = n + 1 → ¬ n0 < m ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 137:
leninv(nσ) = ⌜n1⌝ , nσ[n0] = nσ[n2] + 1, ⌜n0⌝ < # nσ, stutf(nσ), n0 ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[nσ[n2]], nσ[n0] < n1, ⌜n2⌝ < # nσ, n0 < n2, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n0] → ¬ n0 < m, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = nσ[n2] ⊦

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%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2