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Proof step 1 from proof subset-put-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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