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Proof step 21 from proof eq-shift-put (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a0 abort: F discard: F computed: T

Additional information

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