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Proof step 99 from proof pathex-put-nopathex (previous step)

path∃(str, r, r1, st[r0 -- sk, val]) ⊦ path∃(str, r, r1, st), sk = _type, path∃(str, r, r0, st), junkref(str, r, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ validrefs(str, refs, st[r -- _type] .type, st))
Proof step 98:
∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r1 ∧ validrefs(str, refs, (st[r0 -- sk, val][r -- _type]) .type, st[r0 -- sk, val]) ⊦ path∃(str, r, r1, st), sk = _type, path∃(str, r, r0, st), junkref(str, r, st)

Rule argument

[crewrite-arg: pathex-def-simple , F , [st, r0, r, str] → [st[r0 -- sk, val], r1, r, str] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st[r0 -- sk, val], r1, r, str)]]