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Proof step 7 from proof firstsub (previous step)

∀ a0. a0 ∈ a ' + x0 → a0 ∈ y, ∀ y. (∀ a. a ∈ x0 → a ∈ y) → rmdup(x0 + y) = rmdup(y) ⊦ rmdup((a ' + x0) + y) = rmdup(y), a ∈ x0 + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ y. (∀ a. a ∈ x0 → a ∈ y) → rmdup(x0 + y) = rmdup(y), ∀ a0. a0 = a ∨ a0 ∈ x0 → a0 ∈ y ⊦ a ' + rmdup(x0 + y) = rmdup(y), a ∈ x0, a ∈ y

Rule argument

Additional information

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