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Proof step 4 from proof diff-consnotin2 (previous step)

∀ a, x0. x1 \ (a ' + x0) = x1 \ x0 ∨ a ∈ x1 ⊦ (a0 ' + x1) \ (a ' + x0) = (a0 ' + x1) \ x0, a = a0, a ∈ x1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y)
Proof step 6:
¬ ¬ a0 ∈ a ' + x0, ∀ a, x0. x1 \ (a ' + x0) = x1 \ x0 ∨ a ∈ x1 ⊦ (a0 ' + x1) \ (a ' + x0) = (a0 ' + x1) \ x0, a = a0, a ∈ x1
Proof step 8:
¬ a0 ∈ a ' + x0, ∀ a, x0. x1 \ (a ' + x0) = x1 \ x0 ∨ a ∈ x1 ⊦ a0 + (x1 \ (a ' + x0)) = (a0 ' + x1) \ x0, a = a0, a ∈ x1

Rule argument

spec: inst: name: diff-n seq: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y) subst: [y, x, a] → [a ' + x0, x1, a0] flags: rewrite, current

Additional information

Used terms: a ' + x0, x1, a0