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Proof step 20 from proof disjoint-subset (conclusion)

⊦ (∀ r. r ∈ refs → ¬ junkref(str, r, st)) ∧ disjoint(str, refs, st) ∧ st ⊆ st1 → disjoint(str, refs, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
st ⊆ st1, disjoint(str, refs, st), ∀ r. r ∈ refs → ¬ junkref(str, r, st) ⊦ disjoint(str, refs, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]