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Proof step 6 from proof subset-put (previous step)

a0 ∈ st, a ∈ st0, a ∈ st[a, st0[a]], ∀ a0. a0 ∈ st[a, st0[a]] → a0 ∈ st0 ∧ st[a, st0[a]][a0] = st0[a0] ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a0 abort: F discard: F computed: T

Additional information

Used simplifier rules: in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%At-other: ⊦ a ≠ b → st[b, d][a] = st[a]; Used terms: a0