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Proof step 12 from proof union-subset (previous step)

union(asar, n, m0) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m0 → asar[n0] ⊆ as), ∀ n0. n ≤ n0 ∧ ¬ n + m0 < n0 → asar[n0] ⊆ as ⊦ union(asar, n, m0) ⊆ as

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
n0 < n + m0, n ≤ n0, ∀ n0. n ≤ n0 ∧ ¬ n + m0 < n0 → asar[n0] ⊆ as ⊦ m0 = 0, asar[n0] ⊆ as, union(asar, n, m0) ⊆ as

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic