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Proof step 28 from proof diff-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y)
Proof step 30:
¬ ¬ a ∈ z, ∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ (a ∈ x0 ∨ a ∈ y ⊃ rmdup(x0 + y);a ' + rmdup(x0 + y)) \ z = rmdup(((a ' + x0) \ z) + (y \ z))
Proof step 35:
¬ a ∈ z, ∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ (a ∈ x0 ∨ a ∈ y ⊃ rmdup(x0 + y);a ' + rmdup(x0 + y)) \ z = rmdup((a + (x0 \ z)) + (y \ z))

Rule argument

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

Additional information

Used terms: z, x0, a