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Proof step 3 from proof disjoint-in-right (previous step)

a ∈ st1, a ∈ st0, st0 ⊥ st1 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ st0 ⊥ st1 ↔ (∀ a. ¬ a ∈ st0 ∨ ¬ a ∈ st1)
Proof step 5:
a ∈ st0, a ∈ st1, ∀ a. ¬ a ∈ st0 ∨ ¬ a ∈ st1 ⊦

Rule argument

spec: <toplevel> name: disjoint seq: ⊦ st0 ⊥ st1 ↔ (∀ a. ¬ a ∈ st0 ∨ ¬ a ∈ st1) subst: [st1, st0] → [st1, st0] flags: <all paths>

Additional information

Used terms: st1, st0