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Proof step 3 from proof fits-no-refval-01 (conclusion)

⊦ ¬ is_referencevalue(st[r -- sks]) ∧ ¬ junkref(str, r, st) → ¬ fits(r, sks + sk ', st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
fits(r, sks + sk ', st) ⊦ is_referencevalue(st[r -- sks]), junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]