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Proof step 15 from proof subset-put-01 (previous step)

a0 ∈ st, (∀ 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]) ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]

User applied rule case distinction

And obtained the following new goals

Proof step 16:
(∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a]) ∧ ¬ a ∈ st, a0 ∈ st ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]
Proof step 18:
a ∈ st ∧ st[a] = d ∧ (∀ a0. a0 ∈ st -- a → a0 ∈ st0 -- a ∧ st[a0] = st0[a0]), a0 ∈ st ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]

Rule argument

pos: left 2

Proof step 15 from proof subset-put-01 (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information