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Proof step 3 from proof rm-size-in-01 (previous step)

# rmdup(x + y + z) = # rmdup(x + z), a ∈ y ⊦ a ∈ z, a ∈ x

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
# x = # y, ¬ dups(x), ¬ dups(y), ∀ a. a ∈ x → a ∈ y ⊦ ∀ a. a ∈ y → a ∈ x
Proof step 5:
# rmdup(x + y + z) = # rmdup(x + z), a ∈ y, ∀ a. a ∈ rmdup(x + y + z) → a ∈ rmdup(x + z) ⊦ a ∈ x, a ∈ z

Rule argument

spec: <toplevel> name: dups-eqsize-01 seq: # x = # y, ¬ dups(x), ¬ dups(y), ∀ a. a ∈ x → a ∈ y ⊦ ∀ a. a ∈ y → a ∈ x subst: [x, y] → [rmdup(x + z), rmdup(x + y + z)]

Additional information

Used simplifier rules: rm-nodups: ⊦ ¬ dups(rmdup(x));%rm-in: ⊦ a ∈ rmdup(x + y) ↔ a ∈ x ∨ a ∈ y;% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: rmdup(x + z), rmdup(x + y + z)