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Proof step 30 from proof subset-is-union (previous step)

st0 .keys ∩ st .keys = ∅, ∀ rk0. (rk0 ∈ st1[rk, st0[rk]] ↔ rk0 ∈ st ∪ st0) ∧ (st1[rk, st0[rk]][rk0]) = (st ∪ st0)[rk0], 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 = ∅) ⊦ rk0 ∈ st1 ↔ rk0 = rk ∨ rk0 ∈ st0 ∨ rk0 ∈ st, ∃ 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 29:
st0 .keys ∩ st .keys = ∅, 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 = ∅), ∀ rk1. (rk1 ∈ st1 ↔ rk1 ∈ st ∪ st0) ∧ ((rk1 = rk ⊃ st0[rk]; st1[rk1])) = (st ∪ st0)[rk1] ⊦ rk0 ∈ st1 ↔ rk0 = rk ∨ rk0 ∈ st0 ∨ rk0 ∈ st, ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))], [CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st], [CSIMPSEQ : ⊦ rks ∩ rks0 = rks0 ∩ rks], [CSIMPSEQ : ⊦ (rks ∩ rks0) ∩ rks1 = rks ∩ rks0 ∩ rks1])]]