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Proof step 1 from proof diff-app-01 (conclusion)

⊦ x \ (y + x) = []

User applied rule structural induction

And closed the goal!

Rule argument

indvar: x

Additional information

Used simplifier rules: diff-e: ⊦ [] \ y = [];%diff-empty-sub: ⊦ x \ y = [] ↔ x ⊆ y;%subset-app-04: ⊦ x + y ⊆ z + y ↔ x ⊆ z + y;%subset-one: ⊦ a ' ⊆ x ↔ a ∈ x;%list freely generated by [], +;;%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%app: ⊦ x + [] = x; from specification list