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Proof step 46 from proof subset-is-union (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 31:
st0 .keys ∩ st .keys = ∅, st1[rk, st0[rk]] = st ∪ st0, st0 -- rk ⊆ st1 -- rk, rk ∈ st1, st0 ⊆ st1[rk, st0[rk]], rk ∈ st .keys, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ rk0 ∈ st1 ↔ rk0 = rk ∨ rk0 ∈ st0 ∨ rk0 ∈ st, ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅
Proof step 45:
st0 .keys ∩ st .keys = ∅, st1[rk, st0[rk]] = st ∪ st0, st0 -- rk ⊆ st1 -- rk, rk ∈ st1, st0 ⊆ st1[rk, st0[rk]], rk ∈ st .keys, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ st1[rk0] = ((rk0 = rk ⊃ st1[rk]; ((st -- rk) ∪ st0)[rk0])), ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]