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Proof step 76 from proof subset-is-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ st ∪ st0[rk, val] = (st ∪ st0)[rk, val]
Proof step 75:
st0[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) = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ st[rk, val] .keys = st .keys ++ rk])]]