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Proof step 74 from proof subset-is-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk × val ∈ st ↔ rk ∈ st ∧ st[rk] = val])]]