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Proof step 42 from proof diff-union (previous step)

∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ a ' + (rmdup(x0 + y) \ z) = rmdup((a ' + (x0 \ z)) + (y \ z)), a ∈ z, a ∈ x0, a ∈ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 43:
⊦ ¬ a ∈ x → rmdup(a ' + x) = a + rmdup(x)
Proof step 44:
¬ a ∈ (x0 \ z) + (y \ z), ∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ a ' + (rmdup(x0 + y) \ z) = a + rmdup((x0 \ z) + (y \ z)), a ∈ z, a ∈ x0, a ∈ y

Rule argument

spec: list-dup inst: name: rmdup-n seq: ⊦ ¬ a ∈ x → rmdup(a ' + x) = a + rmdup(x) subst: [x, a] → [(x0 \ z) + (y \ z), a] flags: rewrite

Additional information

Used simplifier rules: diff-in: ⊦ a ∈ x \ y ↔ a ∈ x ∧ ¬ a ∈ y;%rmdup-y: ⊦ a ∈ x → rmdup(a ' + x) = rmdup(x); from specification list-dup%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: (x0 \ z) + (y \ z), a