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Proof step 15 from proof diff-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y)
Proof step 17:
¬ ¬ a ∈ z, ∀ z. rmdup(y0) \ z = rmdup(y0 \ z) ⊦ (a ' + rmdup(y0)) \ z = rmdup((a ' + y0) \ z), a ∈ y0
Proof step 19:
¬ a ∈ z, ∀ z. rmdup(y0) \ z = rmdup(y0 \ z) ⊦ (a ' + rmdup(y0)) \ z = rmdup(a + (y0 \ z)), a ∈ y0

Rule argument

spec: inst: name: diff-n seq: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y) subst: [y, x, a] → [z, y0, a] flags: rewrite, current

Additional information

Used terms: z, y0, a