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Proof step 99 from proof ex-fits-validrefs-stronger (conclusion)

¬ junkref(str, r, st), fits(r, sks, st) ⊦ ∃ refs. validrefs(str, r ' + refs, st[r -- _type] .type, st) ∧ # refs = # sks ∧ (∀ sks0. sks0 ⊑ sks ∧ sks0 ≠ [] → st[r -- sks0] .refval = (r ' + refs)[# sks0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 98:
fits(r, sks, st) ⊦ junkref(str, r, st), ∃ refs. validrefs(str, r ' + refs, st[r -- _type] .type, st) ∧ # refs = # sks ∧ (∀ sks0. sks0 ⊑ sks ∧ sks0 ≠ [] → st[r -- sks0] .refval = (r ' + refs)[# sks0])

Rule argument

[emptyarg]

Additional information

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