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Proof step 6 from proof insfirst (previous step)

∀ a, d, ax0, d0. (ax1 + (a × d)' + ax0)[a, d0] = ax1 + (a × d0)' + ax0 ∨ a ∈ ax1 ⊦ (((a0 × d1)' + ax1) + (a × d)' + ax0)[a, d0] = ((a0 × d1)' + ax1) + (a × d0)' + ax0, a ∈ ax1, a ∈ (a0 × d1)'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a, d, ax0, d0. (ax1 + (a × d)' + ax0)[a, d0] = ax1 + (a × d0)' + ax0 ∨ a ∈ ax1 ⊦ (ax1 + (a × d)' + ax0)[a, d0] = ax1 + (a × d0)' + ax0, a = a0, a ∈ ax1

Rule argument

Additional information

Used simplifier rules: inone: ⊦ a ∈ (b × d0)' ↔ a = b;%put-no: ⊦ a ≠ b → ((b × d0)' + ax)[a, d] = (b × d0)' + ax[a, d];% ⊦ ax + ax0 = ax + ax1 ↔ ax0 = ax1; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclist]