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Proof step 5 from proof add-in-02 (previous step)

r = r0 ∨ r ∈ addobj(r0, class1, fis0, st), ∀ r, r0, class1, st. r ∈ addobj(r0, class1, fis0, st) ↔ r = r0 ∨ r ∈ st ⊦ r = r0 ∨ r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r ∈ addobj(r0, class1, fis0, st), ∀ r, r0, class1, st. r ∈ addobj(r0, class1, fis0, st) ↔ r = r0 ∨ r ∈ st ⊦ r = r0, r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]