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Proof step 2 from proof dups-eqsize-ext (previous step)

∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, dups(y), dups(x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
¬ dups(x), ¬ dups(y), ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y
Proof step 4:
¬(¬ dups(y) ∧ ¬ dups(x) ∧ (∀ a. a ∈ y ↔ a ∈ x)), ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, dups(y), dups(x)

Rule argument

spec: inst: name: dups-eqsize seq: ¬ dups(x), ¬ dups(y), ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y subst: [y, x] → [x, y] flags: rewrite, current

Additional information

Used terms: x, y