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Proof step 3 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 4:
¬ dups(x), ¬ dups(y), ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y
Proof step 5:
∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, dups(x), dups(y), ¬ dups(y) ∧ ¬ dups(x) ∧ (∀ a. a ∈ y ↔ a ∈ x)

Rule argument

spec: <toplevel> name: dups-eqsize seq: ¬ dups(x), ¬ dups(y), ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y subst: [y, x] → [x, y] flags: <all paths>

Additional information

Used terms: x, y