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Proof step 14 from proof nodes-no-change-eq (previous step)

m ∈ gstore0 ↔ m ∈ gstore1, gstore0[m] = groupblock(gndx0, lax0, m3), gstore1[m] = groupblock(gndx, lax, m2), m ∈ gstore1 → gndx = gndx0 ∧ lax = lax0 ⊦ ( ¬ m ∈ ginval → (@(m, m0, m1).lnum ∈ gstore0 ∧ @(@(m, m0, m1).pos, @(m, m0, m1).size) ∈ gstore0[@(m, m0, m1).lnum].ladrlist ↔ @(m, m0, m1) ∈ gstore1)) ∧ ( (@(m, m0, m1).lnum ∈ gstore0 ∧ @(@(m, m0, m1).pos, @(m, m0, m1).size) ∈ gstore0[@(m, m0, m1).lnum].ladrlist) ∧ ¬ m ∈ ginval → nodes(gstore0, ginval)[@(m, m0, m1)] = nodes(gstore1, ginval)[@(m, m0, m1)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
m ∈ gstore0 ↔ m ∈ gstore1, gstore0[m] = groupblock(gndx0, lax0, m3), gstore1[m] = groupblock(gndx, lax, m2), m ∈ gstore1 → gndx = gndx0 ∧ lax = lax0 ⊦ (¬ m ∈ ginval → (m ∈ gstore1 ∧ @(m0, m1) ∈ lax0 ↔ @(m, m0, m1) ∈ gstore1)) ∧ (m ∈ gstore1 ∧ @(m0, m1) ∈ lax0 ∧ ¬ m ∈ ginval → nodes(gstore0, ginval)[@(m, m0, m1)] = nodes(gstore1, ginval)[@(m, m0, m1)])

Rule argument

Additional information

Used simplifier rules: ⊦ groupblock(gndx, lax, m).ladrlist = lax; (s)% ⊦ @(m, m0, m1).size = m1; (s)% ⊦ @(m, m0, m1).pos = m0; (s)% ⊦ @(m, m0, m1).lnum = m; (s)