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Proof step 9 from proof junkref-put-nopathex-equiv-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
is_valid_ref(str, r0, st), junkrefs(str, refs, st[refs .first -- _type] .type, st[r0 -- sk, val]) ⊦ sk = _type, refs = [], path∃(str, refs .first, r0, st), ∃ refs0. refs0 ≠ [] ∧ refs0 .first = refs .first ∧ junkrefs(str, refs0, st[refs0 .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

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