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Proof step 24 from proof subset-is-union (previous step)

st1[rk0] = (st ∪ st0)[rk0], rk0 ∈ st1 ↔ rk0 ∈ st ∨ rk0 ∈ st0, st0 .keys ∩ st .keys = ∅, rk ∈ st .keys, st0 ⊆ st1[rk, st0[rk]], rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ rk1. (rk1 ∈ st1 ↔ rk1 ∈ st ∪ st0) ∧ ((rk1 = rk ⊃ st0[rk]; st1[rk1])) = (st ∪ st0)[rk1], ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ rk0 = rk, rk0 ∈ st1 ↔ rk0 ∈ st0 ∨ rk0 ∈ st, ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

User applied rule case distinction

And obtained the following new goals

Proof step 20:
⊦ ¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk]
Proof step 21:
⊦ ∅[rk] = noval
Proof step 22:
⊦ ∅[rk] = noval
Proof step 23:
⊦ ¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk]

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]], [CSIMPSEQ : ⊦ (rks ∩ rks0) ∩ rks1 = rks ∩ rks0 ∩ rks1], [CSIMPSEQ : ⊦ rks ∩ rks0 = rks0 ∩ rks])]]