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Proof step 2 from proof disjoint-nopathex-same (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 2, [extquanttermlist: (r0, r, r0) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ path∃(str, r, r, st) ↔ r = jvmref ∨ r ≠ jvmref ∧ is_valid_ref(str, r, st)], [CSIMPSEQ : ⊦ ¬ is_valid_ref(str, r1, st) ∧ r1 ≠ jvmref → ¬ path∃(str, r, r1, st)])], [EXTRATERMS : (r0, r, r0)]]