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Proof step 7 from proof newrefs-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ newref(r, st) ↔ r ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r)
Proof step 6:
r0 ∈ refs, ∀ r0. r0 ∈ refs → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r0), ∀ rk. rk ∈ st → rk .ref ≠ r ⊦ r = jvmref, r ∈ refs, dups(refs), newref(r0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]