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

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

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

Rule argument

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

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%a: ⊦ (x + y) + z = x + y + z; from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list Used terms: a ', x0, a0