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Proof step 3 from proof disjoint-nopathex-same (previous step)

∀ r3, r2, r1. ¬ (r3 ∈ r ' + r0 ' ∧ r2 ∈ r ' + r0 ' ∧ r3 ≠ r2 ∧ r1 ≠ jvmref ∧ path∃(str, r3, r1, st) ∧ path∃(str, r2, r1, st)), path∃(str, r, r0, st) ⊦ r = r0, r = jvmref, r0 = jvmref

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
path∃(str, r, r0, st), ∀ r1, r2, r3. r1 ≠ r ∧ r1 ≠ r0 ∨ r2 ≠ r ∧ r2 ≠ r0 ∨ r1 = r2 ∨ r3 = jvmref ∨ ¬ path∃(str, r1, r3, st) ∨ ¬ path∃(str, r2, r3, st) ⊦ r = r0, r = jvmref, r0 = jvmref

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ ¬ r ∈ r0 ' + refs ↔ r ≠ r0 ∧ ¬ r ∈ refs])]]