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Proof step 25 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)), valid⊆(str, st, st0) ⊦ valid⊆(str, st, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
valid⊆(str, st, st0), ∀ r. ¬ r ∈ st0 ∧ r ∈ st1 → ¬ junkref(str, r, st1), ∀ r, ty. validref(str, r, ty, st0) → validref(str, r, ty, st1) ⊦ valid⊆(str, st, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]