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Proof step 10 from proof stsubset-new (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
rk ∈ st, ∀ rk. rk ∈ st → rk .ref ≠ r, ∀ rk. rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val] ⊦ r = jvmref, rk ∈ st0 ∧ st[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]