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Proof step 12 from proof add-in-04 (conclusion)

⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ rk ∈ st → rk ∈ addobj(r, class1, fis, st)
Proof step 2:
⊦ r ∈ addobj(r0, class1, fis, st) ↔ r = r0 ∨ r ∈ st
Proof step 11:
r -- sk ∈ addobj(r0, class1, fis, st), r ∈ st ⊦ r = r0, r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])]]