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Proof step 1 from proof subset-put-02 (conclusion)

⊦ st[a, d] ⊆ st0 ↔ st -- a ⊆ st0 -- a ∧ a × d ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∀ a0. a0 ∈ st[a, d] → a0 ∈ st0 ∧ st[a, d][a0] = st0[a0]) ↔ (∀ a0. a0 ∈ st -- a → a0 ∈ st0 -- a ∧ st[a0] = st0[a0]) ∧ a ∈ st0 ∧ st0[a] = d

Rule argument

Additional information

Used simplifier rules: Subset: ⊦ st1 ⊆ st2 ↔ (∀ a. a ∈ st1 → a ∈ st2 ∧ st1[a] = st2[a]);%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a];%in-pair: ⊦ a × d ∈ st ↔ ¬ ¬(a ∈ st ∧ st[a] = d);