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Proof step 11 from proof firstsub (
previous step
)
(∀ a. a ∈ x0 → a ∈ y) → rmdup(x0 + y) = rmdup(y), ∀ a0. a0 ∈ a ' ∨ a0 ∈ x0 → a0 ∈ y, a ∈ x0 ∨ a ∈ y ⊦ rmdup(x0 + y) = rmdup(y)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 12
:
a0 ∈ x0, ∀ a0. a0 ∈ a ' ∨ a0 ∈ x0 → a0 ∈ y, a ∈ x0 ∨ a ∈ y ⊦ rmdup(x0 + y) = rmdup(y), a0 ∈ y
Rule argument
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