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Proof step 20 from proof diff-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ ¬ a ∈ x → rmdup(a ' + x) = a + rmdup(x)
Proof step 22:
¬ a ∈ y0 \ z, ∀ z. rmdup(y0) \ z = rmdup(y0 \ z) ⊦ a ' + (rmdup(y0) \ z) = a + rmdup(y0 \ z), a ∈ y0, a ∈ z

Rule argument

spec: list-dup inst: name: rmdup-n seq: ⊦ ¬ a ∈ x → rmdup(a ' + x) = a + rmdup(x) subst: [x, a] → [y0 \ 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%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: y0 \ z, a