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Proof step 13 from proof junkref-put-norefval (previous step)

∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, (st[r1 -- sk, val][r -- _type]) .type, st[r1 -- sk, val]), is_fskey(sk), okval(val, sk .fs .type) ⊦ sk = _length, sk .fs .type = void_type, ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[r -- _type] .type, st), is_referencevalue(val)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
junkrefs(str, refs, (st[r1 -- sk, val][refs .first -- _type]) .type, st[r1 -- sk, val]), is_fskey(sk), okval(val, sk .fs .type) ⊦ sk = _length, sk .fs .type = void_type, refs = [], is_referencevalue(val), ∃ refs0. refs0 ≠ [] ∧ refs0 .first = refs .first ∧ junkrefs(str, refs0, st[refs .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]