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Proof step 10 from proof junkref-put-nopathex-equiv-01 (previous step)

junkref(str, r, st[r0 -- sk, val]), is_valid_ref(str, r0, st) ⊦ sk = _type, path∃(str, r, r0, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ junkref(str, r, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st))
Proof step 9:
∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, (st[r0 -- sk, val][r -- _type]) .type, st[r0 -- sk, val]), is_valid_ref(str, r0, st) ⊦ sk = _type, path∃(str, r, r0, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

Rule argument

[speclemmaarg: , junkref(str, r, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st)) ,[st, r, str] → [st[r0 -- sk, val], r, str] ,F ,junkrefs , ,junkref-def ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st[r0 -- sk, val], r, str)]]