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Proof step 9 from proof at-put (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ a ≠ b → ((b × d)' + ax)[a] = ax[a]
Proof step 11:
a ≠ a0, ∀ a, b, d. a = b ∨ ax0[b, d][a] = ax0[a] ⊦ b = a0, ax0[b, d][a] = ((a0 × d0)' + ax0)[a], a = b

Rule argument

name: assoc-no rotate: F subst: [a, ax, d, b] → [a, ax0[b, d], d0, a0]

Additional information

Used simplifier rules: assoc-yes: ⊦ ((a × d)' + ax)[a] = d;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; Used terms: a, ax0[b, d], d0, a0