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Proof step 31 from proof disjoint-def-pathex (conclusion)

⊦ disjoint(str, refs, st) ↔ ¬ (∃ r, r0, r1. r ∈ refs ∧ r0 ∈ refs ∧ r ≠ r0 ∧ r1 ≠ jvmref ∧ path∃(str, r, r1, st) ∧ path∃(str, r0, r1, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
⊦ disjoint(str, refs, st) ↔ (∀ r, r0, r1. ¬ r ∈ refs ∨ ¬ r0 ∈ refs ∨ r = r0 ∨ r1 = jvmref ∨ ¬ path∃(str, r, r1, st) ∨ ¬ path∃(str, r0, r1, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]