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Proof step 18 from proof disjoint-subset (previous step)

st ⊆ st1, disjoint(str, refs, st), ∀ r. r ∈ refs → ¬ junkref(str, r, st) ⊦ ∀ r, r0, r1. ¬ (r ∈ refs ∧ r0 ∈ refs ∧ r ≠ r0 ∧ r1 ≠ jvmref ∧ path∃(str, r, r1, st1) ∧ path∃(str, r0, r1, st1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
disjoint(str, refs, st), st ⊆ st1, path∃(str, r0, r, st1), path∃(str, r1, r, st1), r0 ∈ refs, r1 ∈ refs, ∀ r. r ∈ refs → ¬ junkref(str, r, st) ⊦ r1 = r0, r = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]