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Proof step 2 from proof subset-put-strong-04 (previous step)

st ⊆ st0 ⊦ st[a, d][a0, d0] ⊆ st0[a, d][a0, d0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ st1 ⊆ st2 ↔ (∀ a. a ∈ st1 → a ∈ st2 ∧ st1[a] = st2[a])
Proof step 4:
st ⊆ st0 ⊦ ∀ a1. a1 ∈ st[a, d][a0, d0] → a1 ∈ st0[a, d][a0, d0] ∧ st[a, d][a0, d0][a1] = st0[a, d][a0, d0][a1]

Rule argument

spec: store inst: name: Subset seq: ⊦ st1 ⊆ st2 ↔ (∀ a. a ∈ st1 → a ∈ st2 ∧ st1[a] = st2[a]) subst: [st2, st1] → [st0[a, d][a0, d0], st[a, d][a0, d0]] flags: rewrite, current

Additional information

Used terms: st0[a, d][a0, d0], st[a, d][a0, d0]