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Proof step 3 from proof subset-rmdup (previous step)

∀ y. x0 ⊆ y → rmdup(x0 + y) = rmdup(y), a + x0 ⊆ y ⊦ rmdup((a + x0) + y) = rmdup(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ y, x0 ⊆ y, ∀ y. x0 ⊆ y → rmdup(x0 + y) = rmdup(y) ⊦ rmdup(x0 + y) = rmdup(y)

Rule argument

Additional information

Used simplifier rules: subset-rec: ⊦ a ' + x ⊆ y ↔ a ∈ y ∧ x ⊆ y;%rm-in-04: ⊦ a ∈ y → rmdup(a ' + x + y) = rmdup(x + y); from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list