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Proof step 213 from proof pathex-put-nopathex (previous step)

∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r1 ∧ validrefs(str, refs, st[r -- _type] .type, st) ⊦ path∃(str, r, r1, 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 212:
validrefs(str, refs, st[refs .first -- _type] .type, st) ⊦ sk = _type, refs = [], path∃(str, refs .first, refs .last, st[r0 -- sk, val]), path∃(str, refs .first, r0, st), junkref(str, refs .first, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]