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Proof step 5 from proof onenotin-diff (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ¬ a ∈ x → x \ (a ' + x0) = x \ x0
Proof step 7:
∀ a. a ' \ x0 = a ' ∨ a ∈ x0 ⊦ a ' \ x0 = a ', a = a0, a ∈ x0, a0 ∈ a '

Rule argument

spec: <toplevel> name: diff-consnotin2 seq: ⊦ ¬ a ∈ x → x \ (a ' + x0) = x \ x0 subst: [x0, a, x] → [x0, a0, a '] flags: <all paths>

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s) Used terms: x0, a0, a '