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Proof step 10 from proof firstsub (previous step)

∀ y. (∀ 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)

User applied rule all left

And obtained the following new goals

Proof step 11:
(∀ 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)

Rule argument

pos: left 1 terms: y abort: F discard: T computed: T

Additional information

Used terms: y