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Proof step 19 from proof nopathex-update (previous step)

path∃(str, r, r0, update(r0, fis, st)[r0 -- sk, val]), ∀ str0, r1, r2, st0. path∃(str0, r1, r2, update(r2, fis, st0)) → junkref(str0, r1, st0) ∨ path∃(str0, r1, r2, st0) ⊦ _type = sk, _type ∈ fis, junkref(str, r, st), path∃(str, r, r0, st), dups(fis .keys), sk ∈ fis .keys

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ¬ sk ∈ fis → (update(r0, fis, st)[r -- sk, val]) = update(r0, fis, st[r -- sk, val])
Proof step 12:
¬ ¬ sk ∈ fis, path∃(str, r, r0, update(r0, fis, st)[r0 -- sk, val]), ∀ str0, r1, r2, st0. path∃(str0, r1, r2, update(r2, fis, st0)) → junkref(str0, r1, st0) ∨ path∃(str0, r1, r2, st0) ⊦ _type = sk, _type ∈ fis, junkref(str, r, st), path∃(str, r, r0, st), dups(fis .keys), sk ∈ fis .keys
Proof step 18:
¬ sk ∈ fis, path∃(str, r, r0, update(r0, fis, st[r0 -- sk, val])), ∀ str0, r1, r2, st0. path∃(str0, r1, r2, update(r2, fis, st0)) → junkref(str0, r1, st0) ∨ path∃(str0, r1, r2, st0) ⊦ _type = sk, _type ∈ fis, junkref(str, r, st), path∃(str, r, r0, st), dups(fis .keys), sk ∈ fis .keys

Rule argument

[speclemmaarg: , ¬ sk ∈ fis → (update(r0, fis, st)[r -- sk, val]) = update(r0, fis, st[r -- sk, val]) ,[val, sk, r, st, fis, r0] → [val, sk, r0, st, fis, r0] ,F ,okfuns , ,update-put-notin ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val, sk, r0, st, fis, r0)]]