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Proof step 22 from proof validsubset-trans (previous step)

∀ r, ty. validref(str, r, ty, st0) → validref(str, r, ty, st1), ∀ r. ¬ r ∈ st0 ∧ r ∈ st1 → ¬ junkref(str, r, st1), ∀ r. ¬ r ∈ st ∧ r ∈ st0 → ¬ junkref(str, r, st0), ∀ r, ty. validref(str, r, ty, st) → validref(str, r, ty, st0) ⊦ valid⊆(str, st, st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ valid⊆(str, st, st0) ↔ (∀ r, ty. validref(str, r, ty, st) → validref(str, r, ty, st0)) ∧ (∀ r. ¬ r ∈ st ∧ r ∈ st0 → ¬ junkref(str, r, st0))
Proof step 21:
∀ r, ty. validref(str, r, ty, st0) → validref(str, r, ty, st1), ∀ r. ¬ r ∈ st0 ∧ r ∈ st1 → ¬ junkref(str, r, st1), ∀ r. ¬ r ∈ st ∧ r ∈ st0 → ¬ junkref(str, r, st0), ∀ r, ty. validref(str, r, ty, st) → validref(str, r, ty, st0) ⊦ (∀ r, ty. validref(str, r, ty, st) → validref(str, r, ty, st1)) ∧ (∀ r. ¬ r ∈ st ∧ r ∈ st1 → ¬ junkref(str, r, st1))

Rule argument

[crewrite-arg: validsubset-def, F, [st0, st, str] → [st1, st, str]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st1, st, str)]]