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Proof step 44 from proof subset-is-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
st1[rk, st0[rk]] = st ∪ st0, st0 .keys ∩ st .keys = ∅, rk ∈ st .keys, st0 ⊆ st1[rk, st0[rk]], rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ rk0 = rk, st1[rk0] = ((st -- rk) ∪ st0)[rk0], ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rks ∩ rks0 = rks0 ∩ rks], [CSIMPSEQ : ⊦ (rks ∩ rks0) ∩ rks1 = rks ∩ rks0 ∩ rks1])]]