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Proof step 15 from proof stsubset-new (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
rk ∈ st, ∀ rk. rk ∈ st → rk .ref ≠ r, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk] ⊦ rk = r -- sk, r = jvmref, st[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]