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

⊦ (∀ r, r0, class1, st. r ∈ addobj(r0, class1, fis0, st) ↔ r = r0 ∨ r ∈ st) → (r ∈ addobj(r0, class1, (sk × val) ' + fis0, st) ↔ r = r0 ∨ r ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ addobj(r, class1, (sk × val) ' + fis, st) = (addobj(r, class1, fis, st)[r -- sk, val])
Proof step 9:
∀ r, r0, class1, st. r ∈ addobj(r0, class1, fis0, st) ↔ r = r0 ∨ r ∈ st ⊦ r ∈ (addobj(r0, class1, fis0, st)[r0 -- sk, val]) ↔ r = r0 ∨ r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ st[r0 -- sk, val] ↔ r = r0 ∨ r ∈ st], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1])]]