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Proof step 17 from proof newrefs-rec (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 13:
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)
Proof step 16:
r0 = r, ∀ 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)

Rule argument

[CASEDARG : pos: left 4, ()]

Additional information

[[CONCRETESIMPRULES : ()]]