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Proof step 9 from proof add-in-other (conclusion)

⊦ r ≠ r0 ∨ r = r0 ∧ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ r = r0 ∧ is_indexkey(sk) ∧ (sk .index < 0 ∨ n→i(# vals) ≤ sk .index) → (r0 -- sk ∈ addarray(r, ty, vals, st) ↔ r0 -- sk ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ rk ∈ st → rk ∈ addarray(r, ty, vals, st)
Proof step 2:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, vals, st) ↔ r -- sk ∈ st)
Proof step 3:
⊦ r ∈ addarray(r, ty0, vals, st)
Proof step 8:
r0 -- sk ∈ addarray(r0, ty, vals, st), is_indexkey(sk) ∧ (sk .index < 0 ∨ n→i(# vals) ≤ sk .index) ∨ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ⊦ r0 -- sk ∈ st

Rule argument

[emptyarg]

Additional information

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