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Proof step 16 from proof disjoint-subset (previous step)

∀ r, r0, r1. ¬ (r ∈ refs ∧ r0 ∈ refs ∧ r ≠ r0 ∧ r1 ≠ jvmref ∧ path∃(str, r, r1, st) ∧ path∃(str, r0, r1, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
st ⊆ st1, path∃(str, r0, r, st1), path∃(str, r1, r, st1), r0 ∈ refs, r1 ∈ refs, ∀ r. r ∈ refs → ¬ junkref(str, r, st), ∀ r2, r3, r4. ¬ r2 ∈ refs ∨ ¬ r3 ∈ refs ∨ r2 = r3 ∨ r4 = jvmref ∨ ¬ path∃(str, r2, r4, st) ∨ ¬ path∃(str, r3, r4, st) ⊦ r1 = r0, r = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]