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Proof step 7 from proof disjoint-nopath-both (previous step)

disjoint(str, r ' + r0 ', st) ∧ path∃(str, r, r1, st) ∧ path∃(str, r0, r2, st) ∧ r ≠ r0, path∃(str, r0, r2, st), path∃(str, r, r1, st), disjoint(str, r ' + r0 ', st), path∃(str, r1, r2, st) ⊦ r = r0, r1 = jvmref, r2 = jvmref, true ∧ r1 ≠ r2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
disjoint(str, r ' + r0 ', st), path∃(str, r, r2, st), path∃(str, r0, r2, st), is_valid_ref(str, r2, st) ⊦ r = r0, r2 = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ path∃(str, r, r, st) ↔ r = jvmref ∨ r ≠ jvmref ∧ is_valid_ref(str, r, st)])]]