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Proof step 15 from proof disjoint-pathex-forward (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
path∃(str, r4, r2, st), path∃(str, r3, r2, st), path∃(str, r, r1, st), disjoint(str, r ' + r0 ', st) ⊦ r = r0, r2 = jvmref, r4 = r3, r4 ≠ r1 ∧ r4 ≠ r0, r3 ≠ r1 ∧ r3 ≠ r0

Rule argument

[emptyarg]

Additional information

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