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Proof step 4 from proof ins3 (previous step)

a0 ≠ a ⊦ stng[a, d1][b, d0][a0] = stng[b, d0][a, d][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ a0 = a, stng[a, d1][b, d0][a0] = stng[b, d0][a0]

Rule argument

Additional information

Used simplifier rules: At-other: ⊦ a ≠ b → stng[b, d][a] = stng[a];