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Proof step 5 from proof add-in-08 (previous step)

sk0 ≠ sk, ∀ r, sk, class, st. r -- sk ∈ addobj(r, class, fis0, st) ↔ sk ∈ fis0 ∨ sk = _type ∨ r -- sk ∈ st ⊦ r -- sk0 ∈ addobj(r, class, fis0, st) ↔ sk0 ∈ fis0 ∨ sk0 = _type ∨ r -- sk0 ∈ st ∨ sk0 = sk

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ r, sk, class, st. r -- sk ∈ addobj(r, class, fis0, st) ↔ sk ∈ fis0 ∨ sk = _type ∨ r -- sk ∈ st ⊦ sk0 = sk, r -- sk0 ∈ addobj(r, class, fis0, st) ↔ sk0 ∈ fis0 ∨ sk0 = _type ∨ r -- sk0 ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]