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Proof step 3 from proof subset (previous step)

∀ a, s. a ∈ x0 ∧ list2set(x0) ⊆ s → a ∈ s ⊦ a0 ∈ a + x0 ∧ list2set(a + x0) ⊆ s → a0 ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a0 ∈ x0, a ∈ s, list2set(x0) ⊆ s, ∀ a, s. a ∈ x0 ∧ list2set(x0) ⊆ s → a ∈ s ⊦ s = ∅, # s = 0, a0 ∈ s

Rule argument

Additional information

Used simplifier rules: list2set-rec: ⊦ list2set(a ' + x) = list2set(x) ++ a;% ⊦ s ≠ ∅ → # s ≠ 0; (forward)% ⊦ a ∈ s → s ≠ ∅; (forward)% ⊦ s ++ a ⊆ s0 ↔ s ⊆ s0 ∧ a ∈ s0;% ⊦ (x + y) + z = x + y + z;% ⊦ a ∈ b ' ↔ a = b;% ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);% ⊦ a + x = a ' + x;