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Proof step 98 from proof pathex-put-nopathex (previous step)

∃ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 97:
validrefs(str, refs, st[refs .first -- _type] .type, st[r0 -- sk, val]) ⊦ sk = _type, refs = [], path∃(str, refs .first, refs .last, st), path∃(str, refs .first, r0, st), junkref(str, refs .first, st)

Rule argument

[emptyarg]

Additional information

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