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

∀ a1, d1, a2, d2. (a1 × d1 ∈ ax0[a2, d2] ↔ a1 = a2 ∧ d1 = d2) ∨ a1 × d1 ∈ ax0 ⊦ a0 × d0 ∈ ((a1 × d1)' + ax0)[a, d] ↔ a0 × d0 = a × d, a0 × d0 = a1 × d1, a0 × d0 ∈ ax0

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:
a ≠ a1, ∀ a1, d1, a2, d2. (a1 × d1 ∈ ax0[a2, d2] ↔ a1 = a2 ∧ d1 = d2) ∨ a1 × d1 ∈ ax0 ⊦ a0 × d0 ∈ (a1 × d1)' + ax0[a, d] ↔ a0 × d0 = a × d, a0 × d0 = a1 × d1, a0 × d0 ∈ ax0

Rule argument

name: put-no rotate: F subst: [d, a, ax, d0, b] → [d, a, ax0, d1, a1]

Additional information

Used simplifier rules: put-yes: ⊦ ((a × d0)' + ax)[a, d] = (a × d)' + ax;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p ≠ p0 → (p ∈ p0 ' + ax ↔ p ∈ ax);% ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0;% ⊦ ¬ p ∈ p0 ' + ax ↔ p ≠ p0 ∧ ¬ p ∈ ax; Used terms: d, a, ax0, d1, a1