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Proof step 4 from proof Del-at (previous step)

∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0[a0, d] -- b)[a] = st0[a0, d][a], a = b

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a = a0, ∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0[a0, d] -- b)[a] = st0[a0, d][a], a = b
Proof step 8:
∀ b, a. (st0 -- b)[a] = st0[a] ∨ a = b ⊦ (st0[a0, d] -- b)[a] = st0[a0, d][a], a = b, a = a0

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0