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Proof step 15 from proof eq-shift-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
a ∈ st, a ∈ st[a, st0[a]], a ∈ st0, ∀ a0. (a0 ∈ st[a, st0[a]] ↔ a0 ∈ st0) ∧ st[a, st0[a]][a0] = st0[a0] ⊦ a0 = a, (a0 ∈ st ↔ a0 ∈ st0) ∧ st[a0] = st0[a0]

Rule argument

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d;%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;