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Proof step 10 from proof add-in-04 (previous step)

∀ r, sk, r0, class1, st. r -- sk ∈ addobj(r0, class1, fis0, st) ∧ r ∈ st → r = r0 ∨ r -- sk ∈ st ⊦ r -- sk ∈ addobj(r0, class1, fv + fis0, st) ∧ r ∈ st → r = r0 ∨ r -- sk ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ r ∈ addobj(r0, class1, fis, st) ↔ r = r0 ∨ r ∈ st
Proof step 9:
r -- sk ∈ addobj(r0, class1, fv ' + fis0, st), r ∈ st, ∀ r, sk, r0, class1, st. r -- sk ∈ addobj(r0, class1, fis0, st) ∧ r ∈ st → r = r0 ∨ r -- sk ∈ st ⊦ r = r0, r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPSEQ : ⊦ fv + fis = fv ' + fis])]]