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Proof step 7 from proof ls-add-bound (previous step)

∀ n0. n0 ∈ ns → n0 < n ⊦ ∀ n0. n0 ∈ ns → n0 < n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n < n0, n0 ∈ ns, ∀ n0. n0 ∈ ns → n0 < n ⊦ # ns = 0, ns = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]% ⊦ m < n → n ≠ 0; (s)%ne-01: ⊦ n ∈ ns → ns ≠ ∅; (forward) from specification set-basic[nat-set]%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)