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Proof step 9 from proof disjoint-nopath-both (previous step)

¬ (disjoint(str, r1 ' + r2 ', st) ∧ r1 ≠ r2 ∧ r1 ≠ jvmref ∧ r2 ≠ jvmref), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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, disjoint(str, r1 ' + r2 ', st) ∧ r1 ≠ r2

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]