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Proof step 20 from proof junkref-put-nopathexplus-equiv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
junkrefs(str, refs, st[refs .first -- _type] .type, st) ⊦ sk = _type, refs = [], junkref(str, refs .first, st[r0 -- sk, val]), path∃+(str, refs .first, r0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]