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Proof step 41 from proof subset-is-union (previous step)

st0 .keys ∩ st .keys = ∅, rk ∈ st .keys, st0 ⊆ st1[rk, st0[rk]], rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ 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 = rk, st1[rk0] = ((st -- rk) ∪ st0)[rk0], ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

User applied rule all left

And obtained the following new goals

Proof step 40:
st0 .keys ∩ st .keys = ∅, rk ∈ st .keys, st0 ⊆ st1[rk, st0[rk]], rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅), (rk0 ∈ st1 ↔ rk0 ∈ st ∪ st0) ∧ ((rk0 = rk ⊃ st0[rk]; st1[rk0])) = (st ∪ st0)[rk0], ∀ rk1. (rk1 ∈ st1 ↔ rk1 ∈ st ∪ st0) ∧ ((rk1 = rk ⊃ st0[rk]; st1[rk1])) = (st ∪ st0)[rk1] ⊦ rk0 = rk, st1[rk0] = ((st -- rk) ∪ st0)[rk0], ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[EXRARG : pos: left 7, [extquanttermlist: (rk0)abort:F, discard:F, computed:F]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk0)]]