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Proof step 6 from proof inj-put-01 (previous step)

⊦ (∀ a0. (a0 ∈ st[a, d][b, d0] ↔ a0 ∈ st[b, d0]) ∧ st[a, d][b, d0][a0] = st[b, d0][a0]) ↔ a ∈ st ∧ st[a] = d, a = b

User applied rule case distinction

And obtained the following new goals

Proof step 7:
∀ a0. (a0 ∈ st[a, d][b, d0] ↔ a0 ∈ st[b, d0]) ∧ st[a, d][b, d0][a0] = st[b, d0][a0] ⊦ a = b, a ∈ st ∧ st[a] = d
Proof step 8:
st[a] = d, a ∈ st ⊦ a = b, ∀ a0. (a0 ∈ st[a, d][b, d0] ↔ a0 ∈ st[b, d0]) ∧ st[a, d][b, d0][a0] = st[b, d0][a0]

Rule argument

pos: right 1

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