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Proof step 16 from proof put-neq (conclusion)

⊦ st ≠ st0 ∧ ¬ a ∈ st ∧ ¬ a ∈ st0 → st[a, d] ≠ st0[a, d]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(Extension,,, ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]))))]