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Proof step 14 from proof junkref-put-norefval (previous step)

junkref(str, r, st[r1 -- sk, val]), is_fskey(sk), okval(val, sk .fs .type) ⊦ sk = _length, sk .fs .type = void_type, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st), is_referencevalue(val)

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 13:
∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, (st[r1 -- sk, val][r -- _type]) .type, st[r1 -- sk, val]), is_fskey(sk), okval(val, sk .fs .type) ⊦ sk = _length, sk .fs .type = void_type, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st), is_referencevalue(val)

Rule argument

[crewrite-arg: junkref-def , F , [st, r, str] → [st[r1 -- sk, val], r, str] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st[r1 -- sk, val], r, str)]]