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Proof step 14 from proof subset-put-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
a0 ∈ st, (∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a]) ∧ ¬ a ∈ st ∨ a ∈ st ∧ st[a] = d ∧ (∀ a0. a0 ∈ st -- a → a0 ∈ st0 -- a ∧ st[a0] = st0[a0]) ⊦ a0 = a, a0 ∈ st0 ∧ 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];