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Proof step 17 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, sk ∈ fis, sk ∈ fis .keys, dups(fis .keys), path∃(str, r, r0, st), junkref(str, r, st), _type ∈ fis

User applied rule all left

And obtained the following new goals

Proof step 16:
path∃(str, r, r0, update(r0, fis, st[r0 -- sk, val])), path∃(str, r, r0, update(r0, fis, st[r0 -- sk, val])) → junkref(str, r, st[r0 -- sk, val]) ∨ path∃(str, r, r0, st[r0 -- sk, val]) ⊦ _type = sk, sk ∈ fis, sk ∈ fis .keys, dups(fis .keys), path∃(str, r, r0, st), junkref(str, r, st), _type ∈ fis

Rule argument

[EXRARG : pos: left 2, [extquanttermlist: (str, r, r0, st[r0 -- sk, val]) abort:F , discard:T , computed:T ]]

Additional information

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