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Proof step 10 from proof cons-03 (previous step)

disj(x, y) ⊦ disj(x + a ', y), a ∈ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ disj(x, y) ↔ (∀ a. ¬(a ∈ x ∧ a ∈ y))

Rule argument

spec: <toplevel> name: disjoint seq: ⊦ disj(x, y) ↔ (∀ a. ¬(a ∈ x ∧ a ∈ y)) subst: [y, x] → [y, x + a '] flags: <all paths>

Additional information

Used simplifier rules: in: ⊦ disj(x, y) ∧ a ∈ x → ¬ a ∈ y;% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: y, x + a '