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Proof step 20 from proof stsubset-new (previous step)

∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
rk ∈ st, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk .ref ≠ r ⊦ r = jvmref, st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ (rk = r -- sk ∨ rk ∈ st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st])]]