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Proof step 5 from proof firstsub (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ x → rmdup(a ' + x) = rmdup(x)
Proof step 7:
∀ 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
Proof step 9:
a ∈ x0 + y, ∀ a0. a0 ∈ a ' + x0 → a0 ∈ y, ∀ y. (∀ a. a ∈ x0 → a ∈ y) → rmdup(x0 + y) = rmdup(y) ⊦ rmdup(x0 + y) = rmdup(y)

Rule argument

spec: list-dup name: rmdup-y seq: ⊦ a ∈ x → rmdup(a ' + x) = rmdup(x) subst: [x, a] → [x0 + y, a] flags: <all paths>

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x0 + y, a