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Proof step 44 from proof fits-in (previous step)

sks0 ≠ [], fits(r, sk ' + sks0, st), ∀ str, r, st. sks0 ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sks0, st) → r -- sks0 ∈ st ⊦ junkref(str, r, st), r -- sk ∈ st ∧ st[r -- sk] .refval -- sks0 ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
fits(r, sk ' + sks0, st), ∀ str0, r0, st0. ¬ junkref(str0, r0, st0) ∧ fits(r0, sks0, st0) → r0 -- sks0 ∈ st0 ⊦ sks0 = [], junkref(str, r, st), r -- sk ∈ st ∧ st[r -- sk] .refval -- sks0 ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]