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Proof step 21 from proof stsubset-new (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 10:
∀ 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
Proof step 20:
∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk .ref ≠ r ⊦ ∀ rk. rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val], r = jvmref

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]