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Proof step 10 from proof is_valid_ref-put-01 (conclusion)

⊦ r ≠ r0 → (is_valid_ref(str, r0, st[r -- sk, val]) ↔ is_valid_ref(str, r0, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ is_valid_ref(str, r, st) ↔ ¬ ¬ (is_valid_ref(r, st) ∧ (is_obj(str, r, st) ∨ is_array(str, r, st)))
Proof step 2:
⊦ r ≠ r0 → (is_valid_ref(r0, st[r -- sk, val]) ↔ is_valid_ref(r0, st))
Proof step 3:
⊦ is_obj(str, r, st) ↔ class+∃(str, st[r -- _type] .type .class) ∧ ¬ is_abstract(st[r -- _type] .type .class) ∧ is_obj(r, st)
Proof step 4:
⊦ r ≠ r0 → (is_obj(r0, st[r -- sk, val]) ↔ is_obj(r0, st))
Proof step 5:
⊦ is_valid_ref(str, r, st) ↔ is_obj(str, r, st) ∨ is_array(str, r, st)
Proof step 6:
⊦ is_obj(r, st) → is_valid_ref(r, st)
Proof step 9:
class+∃(str, st[r0 -- _type] .type .class), is_obj(r0, st) ⊦ r = r0, is_obj(str, r0, st[r -- sk, val]), is_array(str, r0, st), is_abstract(st[r0 -- _type] .type .class)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ≠ r0 → ( is_array (str, r0, st[r -- sk, val]) ↔ is_array(str, r0, st))], [CSIMPSEQ : ⊦ r ≠ r0 → (st[r0 -- sk0, val][r -- sk]) = st[r -- sk]])]]