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Proof step 9 from proof subset-rec (previous step)

a ∈ a ' + x → a ∈ y, ∀ a0. a0 ∈ a ' + x → a0 ∈ y ⊦ a ∈ y ∧ (a0 ∈ x → a0 ∈ y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a ∈ y, a0 ∈ x, ∀ a1. a1 ∈ x → a1 ∈ y ⊦ a0 ∈ y

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)