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Proof step 4 from proof put-neq (previous step)

∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0[a, d0]) ∧ st[a, d][a0] = st0[a, d0][a0] ⊦ d = d0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0[a, d0]) ∧ st[a, d][a0] = st0[a, d0][a0] ⊦ d = d0

Rule argument

Additional information