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Proof step 8 from proof lookup-02 (previous step)

unique_al(ax0), a0 × d0 ∈ ax0, ∀ a0, d0. a0 × d0 ∈ ax0 → ax0[a0] = d0 ⊦ a0 × d0 = a × d, ((a × d)' + ax0)[a0] = d0, a ∈ ax0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ≠ b → ((b × d)' + ax)[a] = ax[a]
Proof step 10:
a0 ≠ a, unique_al(ax0), a0 × d0 ∈ ax0, ∀ a0, d0. a0 × d0 ∈ ax0 → ax0[a0] = d0 ⊦ a0 × d0 = a × d, ax0[a0] = d0, a ∈ ax0

Rule argument

name: assoc-no rotate: F subst: [a, ax, d, b] → [a0, ax0, d, a]

Additional information

Used simplifier rules: assoc-yes: ⊦ ((a × d)' + ax)[a] = d;%in-01: ⊦ a × d ∈ ax → a ∈ ax;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; Used terms: a0, ax0, d, a