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Proof step 10 from proof addobj-in-02 (previous step)

∀ r, r0, sk, class, st. r ≠ r0 ∨ ¬ sk ∈ sks0 ∧ sk ≠ _type → (r -- sk ∈ addobj(r0, class, sks0, st) ↔ r -- sk ∈ st), r ≠ r0 ∨ ¬ sk ∈ sk0 + sks0 ∧ sk ≠ _type ⊦ r -- sk ∈ addobj(r0, class, sk0 + sks0, st) ↔ r -- sk ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ r, r0, sk, class, st. r ≠ r0 ∨ ¬ sk ∈ sks0 ∧ sk ≠ _type → (r -- sk ∈ addobj(r0, class, sks0, st) ↔ r -- sk ∈ st), r ≠ r0 ∨ sk ≠ sk0 ∧ ¬ sk ∈ sks0 ∧ sk ≠ _type ⊦ r -- sk ∈ addobj(r0, class, sk0 ' + sks0, st) ↔ r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ ¬ sk ∈ sk0 ' + sks ↔ sk ≠ sk0 ∧ ¬ sk ∈ sks], [CSIMPSEQ : ⊦ sk + sks = sk ' + sks])]]