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Proof step 13 from proof newrefs-rec (previous step)

r0 ∈ refs, ∀ rk. rk ∈ st → rk .ref ≠ r, r0 ∈ refs → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r0), ∀ r0. r0 ∈ refs → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r0) ⊦ r = jvmref, dups(refs), r ∈ refs, r0 ≠ jvmref ∧ (r0 -- sk ∈ st → r1 ≠ r0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
⊦ r -- sk ∈ st → r ∈ st
Proof step 12:
r0 ∈ st, r0 -- sk ∈ st, r0 ∈ refs, ∀ rk. rk ∈ st → rk .ref ≠ r0, ∀ rk. rk ∈ st → rk .ref ≠ r, ∀ r0. r0 ∈ refs → r0 ≠ jvmref ∧ (∀ rk. rk ∈ st → rk .ref ≠ r0) ⊦ r0 = jvmref, r = jvmref, r ∈ refs, dups(refs)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]