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Proof step 6 from proof at-put (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ≠ b → ((b × d0)' + ax)[a, d] = (b × d0)' + ax[a, d]
Proof step 8:
b ≠ a0, ∀ a, b, d. a = b ∨ ax0[b, d][a] = ax0[a] ⊦ a = b, ((a0 × d0)' + ax0[b, d])[a] = ((a0 × d0)' + ax0)[a]

Rule argument

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

Additional information

Used simplifier rules: put-yes: ⊦ ((a × d0)' + ax)[a, d] = (a × d)' + ax;%assoc-no: ⊦ a ≠ b → ((b × d)' + ax)[a] = ax[a];% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; Used terms: d, b, ax0, d0, a0