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Proof step 16 from proof stsubset-new (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 13:
rk = r -- sk, rk ∈ st, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk .ref ≠ r ⊦ st[rk] = val, r = jvmref
Proof step 15:
rk ≠ r -- sk, rk ∈ st, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk .ref ≠ r ⊦ st[rk] = st0[rk], r = jvmref

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]