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Proof step 77 from proof subset-is-union (previous step)

st ⊆ st1 ⊦ ∃ st0. st1 = st0 ∪ st ∧ st .keys ∩ st0 .keys = ∅

User applied rule structural induction

And obtained the following new goals

Proof step 2:
⊦ st ∪ ∅ = st
Proof step 76:
∀ 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 = ∅

Rule argument

[VARWITHVARSEQSARG : st, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : store generated by ∅, [];], [CSIMPSEQ : ⊦ ∅ ⊆ st], [CSIMPSEQ : ⊦ ∅ .keys = ∅], [CSIMPSEQ : ⊦ rks ∩ ∅ = ∅], [CSIMPSEQ : ⊦ (rks ∩ rks0) ∩ rks1 = rks ∩ rks0 ∩ rks1], [CSIMPSEQ : ⊦ rks ∩ rks0 = rks0 ∩ rks])]]