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Proof step 28 from proof junkref-subset (previous step)

⊦ ¬ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st)) ∧ st ⊆ st1 → ¬ junkref(str, r, st1)

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 27:
⊦ ¬ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st)) ∧ st ⊆ st1 → ¬ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[r -- _type] .type, st1))

Rule argument

[crewrite-arg: junkref-def, F, [st, r, str] → [st1, r, str]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st1, r, str)]]