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Proof step 13 from proof eq-shift-del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
a ∈ st0, ∀ a0. (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0]) ⊦ a0 = a, a ∈ st, (a0 ∈ st ↔ a0 ∈ st0) ∧ (a0 ∈ st → st[a0] = st0[a0])

Rule argument

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];