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Proof step 68 from proof subset-is-union (previous step)

rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ ∀ rk0. rk0 ∈ st0 → rk0 ∈ st1[rk, st0[rk]] ∧ st0[rk0] = (st1[rk, st0[rk]][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 67:
rk0 ∈ st0, 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) = ∅, rk0 ∈ st1 ∧ st0[rk0] = ((rk0 = rk ⊃ st0[rk]; st1[rk0]))

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])]]