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Proof step 21 from proof newrefs-rec (previous step)

¬ r ∈ refs ∧ newref(r, st) ∧ ¬ dups(refs) ∧ (∀ r. r ∈ refs → newref(r, st)) ⊦ ¬ dups(refs) ∧ ¬ r ∈ refs ∧ (∀ r0. r0 ∈ refs ∨ r0 = r → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ newref(r, st) ↔ r ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r)
Proof step 20:
∀ rk. rk ∈ st → rk .ref ≠ r, ∀ r0. r0 ∈ refs → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r0), r0 ∈ refs ∨ r0 = r ⊦ r = jvmref, dups(refs), r ∈ refs, r0 ≠ jvmref ∧ (rk ∈ st → rk .ref ≠ r0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]