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Proof step 12 from proof subset-is-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
st1[rk, st0[rk]] = st ∪ st0, st0 .keys ∩ st .keys = ∅, st0 ⊆ st1[rk, st0[rk]], rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ rk ∈ st, ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅, st1 = st1[rk, st1[rk]] ∧ ¬ rk ∈ st .keys

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (st[rk, val][rk, val0]) = st[rk, val0]], [CSIMPSEQ : ⊦ rks ∩ (rks0 ++ rk) = ∅ ↔ rks ∩ rks0 = ∅ ∧ ¬ rk ∈ rks], [CSIMPSEQ : ⊦ rks ∩ rks0 = rks0 ∩ rks], [CSIMPSEQ : ⊦ (rks ∩ rks0) ∩ rks1 = rks ∩ rks0 ∩ rks1], [CSIMPSEQ : ⊦ ¬ rk ∈ st → st -- rk = st])]]