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Proof step 70 from proof union-addarray-rew (previous step)

r0 ≠ r0 ∨ r0 = r0 ∧ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ r0 = r0 ∧ is_indexkey(sk) ∧ (sk .index < 0 ∨ n→i(# vals) ≤ sk .index), newref(r0, st), newref(r0, st0), is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ is_indexkey(sk) ∧ (sk .index < 0 ∨ n→i(# vals) ≤ sk .index) ⊦ r0 = jvmref, (r0 -- sk ∈ st ∪ st0 ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r0, ty, vals, st0)) ∧ (st ∪ st0)[r0 -- sk] = (st ∪ addarray(r0, ty, vals, st0))[r0 -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
⊦ rk ∈ st → rk ∈ addarray(r, ty, vals, st)
Proof step 69:
newref(r0, st), newref(r0, st0), is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ is_indexkey(sk) ∧ (sk .index < 0 ∨ n→i(# vals) ≤ sk .index) ⊦ r0 = jvmref, (r0 -- sk ∈ addarray(r0, ty, vals, st0) → r0 -- sk ∈ st ∨ r0 -- sk ∈ st0) ∧ (st ∪ st0)[r0 -- sk] = (st ∪ addarray(r0, ty, vals, st0))[r0 -- sk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st ∪ st0 ↔ rk ∈ st ∨ rk ∈ st0])]]