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Proof step 63 from proof subset-is-union (previous step)

∀ rk0. rk0 ∈ st0 -- rk → rk0 ∈ st1 -- rk ∧ (st0 -- rk)[rk0] = (st1 -- rk)[rk0], rk ∈ st1, rk0 ∈ st0, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ rk0 = rk, st0[rk0] = st1[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 62:
rk ∈ st1, rk0 ∈ st0, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅), ∀ rk1. rk1 ∈ st0 -- rk → rk1 ≠ rk ∧ rk1 ∈ st1 ∧ st0[rk1] = st1[rk1] ⊦ rk0 = rk, st0[rk0] = st1[rk0], ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ≠ rk0 → (st -- rk0)[rk] = st[rk]], [CSIMPSEQ : ⊦ rk ∈ st -- rk0 ↔ rk ≠ rk0 ∧ rk ∈ st])]]