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Proof step 23 from proof stsubset-new (previous step)

∀ rk. rk ∈ st → rk .ref ≠ r ⊦ (∀ rk. rk ∈ st → rk ∈ st0[r -- sk, val] ∧ st[rk] = (st0[r -- sk, val][rk])) ↔ st ⊆ st0, r = jvmref

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
∀ rk. rk ∈ st → rk .ref ≠ r ⊦ r = jvmref, (∀ rk. rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val]) ↔ st ⊆ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))])]]