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Proof step 4 from proof subset-delete-01 (previous step)

∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a] ⊦ st -- a ⊆ st0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ st1 ⊆ st2 ↔ (∀ a. a ∈ st1 → a ∈ st2 ∧ st1[a] = st2[a])
Proof step 6:
∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a] ⊦ ∀ a0. a0 ∈ st -- a → a0 ∈ st0 ∧ (st -- a)[a0] = st0[a0]

Rule argument

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

Additional information

Used terms: st0, st -- a