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Proof step 8 from proof stutf-monle (previous step)

nσ[0] = 0, ⌜n⌝ + ⌜1⌝ < # nσ, nσ[n + 1] < nσ[n], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1 ⊦ nσ[n] = 0, nσ = []

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: n

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: n