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Proof step 5 from proof ex-fits-validrefs (previous step)

sks ≠ [], # refs = # sks, fits(r, sks, st), validrefs(str, r ' + refs, st[r -- _type] .type, st), sks ≠ [] → st[r -- sks] .refval = (r ' + refs)[# sks] ⊦ st[r -- sks] .refval = (r ' + refs) .last, junkref(str, r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# refs = # sks, st[r -- sks] .refval = (r ' + refs)[# sks], fits(r, sks, st), validrefs(str, r ' + refs, st[r -- _type] .type, st) ⊦ (r ' + refs)[# sks] = (r ' + refs) .last, sks = [], junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]