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Proof step 16 from proof pathex-put-last (previous step)

r ≠ r0, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str, refs, (st[r0 -- sk, val][r -- _type]) .type, st[r0 -- sk, val]) ⊦ sk = _type, path∃(str, r, r0, st), junkref(str, r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
validrefs(str, refs, st[refs .first -- _type] .type, st[refs .last -- sk, val]) ⊦ sk = _type, refs = [], refs .first = refs .last, path∃(str, refs .first, refs .last, st), junkref(str, refs .first, st), dups(refs)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ≠ sk0 → (st[r -- sk, val][r0 -- sk0]) = st[r0 -- sk0]])]]